Consider the following implications relating to functional and multivalued…

Concept: Functional dependencies (FDs) and multivalued dependencies (MVDs) follow a fixed set of inference axioms. Three of them settle this question: (1) Union/Decomposition for FDs — A → B and A → C combine into A → BC, and conversely A → BC splits into A → B and A → C. (2) Replication — every FD is also an MVD, so A → X always gives A ↠ X. (3) The reverse direction fails in general — an MVD does NOT imply the corresponding FD, so MVD-only premises can never be combined into a fresh FD conclusion.

Application — checking each of the four implications:

1. i. A ↠ B and A ↠ C ⇒ A → BC — not valid. Both premises are MVDs, and by axiom (3) an MVD-only combination can never yield an FD conclusion, no matter how the two MVDs are paired.

2. ii. A → B and A → C ⇒ A ↠ BC — valid. Union (axiom 1) turns the two FDs into A → BC, and replication (axiom 2) then turns that FD into the MVD A ↠ BC.

3. iii. A ↠ BC and A → B ⇒ A → C — not valid. An MVD on BC together with an FD on only part of it (B) does not force an FD on the remaining part (C): a fixed A can still pair with more than one C value even after B is pinned down.

4. iv. A → BC and A → B ⇒ A ↠ C — valid. Decomposition (axiom 1) pulls A → C out of A → BC, and replication (axiom 2) then turns that FD into the MVD A ↠ C.

Cross-check — a small relation for the two rejected implications:

• For implication i: take R(A, B, C) with A = a and, for that single A value, every combination of B ∈ {b1, b2} with C ∈ {c1, c2} present (a full cross-product of four tuples). This satisfies A ↠ B and A ↠ C, but A does not determine BC as one pair — B and C vary independently for the same A, so A → BC fails.

• For implication iii: take R(A, B, C, D) with A = a, B always equal to b0 (so A → B genuinely holds), but C taking two different values (c1, c2) each paired with every D value that occurs for A = a (so A ↠ BC holds by the cross-product test). Since C still varies for the same A, A → C fails even though A ↠ BC and A → B both hold.

Result: exactly two of the four implications (ii and iv) are valid derivations, so the count is 2.