From the following instance of the relation scheme R(A, B, C), we can conclude…

Concept

A functional dependency X → Y holds on a relation only if every pair of tuples that agree on X also agrees on Y. A single instance has an asymmetric power over an FD: it can DISPROVE a dependency (two tuples that match on the left but differ on the right are a counterexample that holds for the whole scheme), but it can never PROVE one (no violation in one instance does not guarantee the dependency holds for every legal instance of the scheme).

Applying it to the instance

1. Test B → C: the first two tuples have B = 1 but C = 1 and C = 0. Two tuples agree on B yet differ on C, which is a direct counterexample. So B → C is violated, and we can firmly conclude B does not functionally determine C.

2. Test A → B: tuples with A = 1 all have B = 1, and tuples with A = 2 all have B = 3, so this instance shows no counterexample to A → B. But by the concept above, the absence of a counterexample in one instance cannot establish that A → B holds for the scheme.

3. Test A → C and the rest similarly: A = 1 maps to both C = 1 and C = 0, so A → C is also disproved; and no left→right claim about A can be asserted as holding from a single instance.

Cross-check / contrast

Only a DISPROVING conclusion survives, because only it is logically valid from one instance:

• “B does not functionally determine C” — a valid conclusion, backed by the B = 1 → {C = 1, C = 0} counterexample.

• Any statement asserting that A functionally determines B (or C) — not concludable, since one violation-free instance never proves an FD for the scheme.

• Any statement asserting that A does NOT functionally determine B — not supported, because this instance shows no counterexample to A → B at all.

Hence the only valid conclusion is: B does not functionally determine C.