Understanding Functional Dependency Properties

We analyze each option based on the properties of functional dependencies (FDs) in relational databases. Let the relation schema have attributes $P, Q, R, S$. $X \rightarrow Y$ denotes a functional dependency, where $X, Y \subseteq \{P, Q, R, S\}$.

Option 1 Analysis: $(\{P, Q\} \rightarrow \{R\} \text{ and } \{P\} \rightarrow \{R\}) \implies \{Q\} \rightarrow \{R\}$

This statement is false. We can demonstrate this with a counterexample.

Consider a relation instance with the following tuples:

• $(p1, q1, r1)$
• $(p1, q2, r1)$
• $(p2, q1, r2)$

Let's check the conditions:

• Check $\{P\} \rightarrow \{R\}$: The value $p1$ maps only to $r1$. The value $p2$ maps only to $r2$. So, $\{P\} \rightarrow \{R\}$ holds.
• Check $\{P, Q\} \rightarrow \{R\}$: The combination $(p1, q1)$ maps to $r1$. The combination $(p1, q2)$ maps to $r1$. The combination $(p2, q1)$ maps to $r2$. Each combination of $P$ and $Q$ maps to a single value of $R$. So, $\{P, Q\} \rightarrow \{R\}$ holds.

Now, let's check the conclusion:

• Check $\{Q\} \rightarrow \{R\}$: The value $q1$ appears with $p1$ (mapping to $r1$) and with $p2$ (mapping to $r2$). Since $q1$ maps to different values ($r1$ and $r2$), $\{Q\} \rightarrow \{R\}$ does not hold.

Since the premise holds but the conclusion does not, this option is not always true.

Option 2 Analysis: $\{P, Q\} \rightarrow \{R\} \implies (\{P\} \rightarrow \{R\} \text{ or } \{Q\} \rightarrow \{R\})$

This statement is false. A counterexample requires $\{P, Q\} \rightarrow \{R\}$ to be true, while *both* $\{P\} \rightarrow \{R\}$ and $\{Q\} \rightarrow \{R\}$ are false.

Consider the following relation instance:

• $(p1, q1, r1)$
• $(p1, q2, r2)$
• $(p2, q1, r3)$
• $(p2, q2, r4)$

Let's check the conditions:

• Check $\{P, Q\} \rightarrow \{R\}$: Each unique pair of $(P, Q)$ values corresponds to a unique $R$ value. This FD holds.
• Check $\{P\} \rightarrow \{R\}$: $P$ value $p1$ corresponds to $R$ values $r1$ and $r2$. This FD does not hold.
• Check $\{Q\} \rightarrow \{R\}$: $Q$ value $q1$ corresponds to $R$ values $r1$ and $r3$. This FD does not hold.

Since $\{P\} \rightarrow \{R\}$ is false and $\{Q\} \rightarrow \{R\}$ is false, the conclusion $(\{P\} \rightarrow \{R\} \text{ or } \{Q\} \rightarrow \{R\})$ is false. The premise is true, but the conclusion is false, so this option is not always true.

Option 3 Analysis: $(\{P\} \rightarrow \{R\} \text{ and } \{Q\} \rightarrow \{S\}) \implies \{P, Q\} \rightarrow \{R, S\}$

This statement is true. This is a standard property derived from Armstrong's axioms, often proven using the augmentation and union rules.

Derivation:

1. $\{P\} \rightarrow \{R\}$ (Given)
2. $\{Q\} \rightarrow \{S\}$ (Given)
3. From (1), using the rule $X \rightarrow Y \implies XZ \rightarrow Y$, we get $\{P, Q\} \rightarrow \{R\}$.
4. From (2), using the rule $X \rightarrow Y \implies XZ \rightarrow Y$, we get $\{P, Q\} \rightarrow \{S\}$.
5. Now we have $\{P, Q\} \rightarrow \{R\}$ and $\{P, Q\} \rightarrow \{S\}$.
6. Using the Union rule (If $X \rightarrow Y$ and $X \rightarrow Z$, then $X \rightarrow YZ$), we combine the results from step 3 and 4: $\{P, Q\} \rightarrow \{R, S\}$.

Therefore, this option is always true.

Option 4 Analysis: $\{P\} \rightarrow \{R\} \implies \{P, Q\} \rightarrow \{R\}$

This statement is true. This property follows directly from the definition of functional dependency and can be proven using Armstrong's axioms.

Proof by contradiction:

1. Assume $\{P\} \rightarrow \{R\}$ holds.
2. Assume $\{P, Q\} \rightarrow \{R\}$ does not hold.
3. If $\{P, Q\} \rightarrow \{R\}$ does not hold, there exist two tuples $t_1$ and $t_2$ such that $t_1.P = t_2.P$ and $t_1.Q = t_2.Q$, but $t_1.R \neq t_2.R$.
4. Let $t_1.P = t_2.P = p$.
5. From the initial assumption (1), since $t_1.P = t_2.P$, it must be that $t_1.R = t_2.R$.
6. This contradicts the condition $t_1.R \neq t_2.R$ derived from the assumption that $\{P, Q\} \rightarrow \{R\}$ does not hold.
7. Therefore, the assumption (2) that $\{P, Q\} \rightarrow \{R\}$ does not hold must be false. Hence, $\{P, Q\} \rightarrow \{R\}$ must be true if $\{P\} \rightarrow \{R\}$ is true.

This demonstrates the Augmentation property where adding attributes to the determinant does not invalidate a dependency if the original determinant alone determines the dependent attributes.

Therefore, this option is always true.

Conclusion

Based on the analysis, options 3 and 4 are the statements that are always true regarding functional dependencies.