Question

Let 𝑃, 𝑄, 𝑅 and 𝑆 be the attributes of a relation in a relational schema. Let 𝑋 ⟶𝑌
indicate functional dependency in the context of a relational database, where
𝑋, 𝑌 ⊆{𝑃, 𝑄, 𝑅, 𝑆}.
Which of the following options is/are always true?

• A. If ( {𝑃, 𝑄} ⟶{𝑅} and {𝑃} ⟶{𝑅} ), then {𝑄} ⟶{𝑅}
• B. If {𝑃, 𝑄} ⟶{𝑅}, then ( {𝑃} ⟶{𝑅} or {𝑄} ⟶{𝑅} )
• C. If ( {𝑃} ⟶{𝑅} and {𝑄} ⟶{𝑆} ), then {𝑃, 𝑄} ⟶{𝑅, 𝑆}
• D. If {𝑃} ⟶{𝑅}, then {𝑃, 𝑄} ⟶{𝑅}

Hint:

Use Armstrong's axioms: augmentation always lets you add extra attributes to the left side of a dependency (option D), and combining two independent dependencies via augmentation plus transitivity gives their union (option C); the reverse-direction splits in options A and B are not generally valid.

Answer 1

The Correct Option is C, D

Step 1: Evaluate option A: if \(\{P,Q\} \rightarrow \{R\}\) and \(\{P\} \rightarrow \{R\}\), then \(\{Q\} \rightarrow \{R\}\). This does not follow from Armstrong's axioms; a larger determinant implying R, together with a smaller subset of it also implying R, gives no information about whether the remaining attribute Q alone determines R. A relation where P alone forces R while Q has no functional relation to R at all satisfies both premises yet violates the conclusion. So option A is FALSE.

Step 2: Evaluate option B: if \(\{P,Q\} \rightarrow \{R\}\), then \(\{P\} \rightarrow \{R\}\) or \(\{Q\} \rightarrow \{R\}\). Functional dependencies do not obey a decomposition rule on the determinant side. A relation where R is computed jointly from both P and Q, with neither alone determining R, satisfies the premise while violating both parts of the conclusion. So option B is FALSE.

Step 3: Evaluate option C: if \(\{P\} \rightarrow \{R\}\) and \(\{Q\} \rightarrow \{S\}\), then \(\{P,Q\} \rightarrow \{R,S\}\). Using Armstrong's axioms: augment \(P \rightarrow R\) with Q to get \(PQ \rightarrow RQ\); augment \(Q \rightarrow S\) with R to get \(RQ \rightarrow RS\). Chaining by transitivity, \(PQ \rightarrow RQ \rightarrow RS\), so \(PQ \rightarrow RS\). This is a provable consequence of the axioms, so option C is always TRUE.

Step 4: Evaluate option D: if \(\{P\} \rightarrow \{R\}\), then \(\{P,Q\} \rightarrow \{R\}\). This is a direct application of the Augmentation axiom: from \(P \rightarrow R\), adding Q to both sides gives \(PQ \rightarrow RQ\), and since \(\{R,Q\} \rightarrow \{R\}\) trivially, transitivity gives \(\{P,Q\} \rightarrow \{R\}\). Option D is always TRUE.

Step 5: The always-true statements are C and D.

\[ \boxed{\text{Options C and D are always true}} \]