Consider a relation R with five attributes V, W, X, Y, and Z. The following…

Compute the closure of VXY:

• Start: {V, X, Y}

• From VY → W, add W: now {V, X, Y, W}

• From WX → Z (we have W and X), add Z: now {V, X, Y, W, Z} — all attributes are obtained.

So (V X Y)+ = {V, W, X, Y, Z} — VXY determines every attribute.

Check minimality by testing proper subsets:

• VY+ = {V, Y, W} (from VY → W). No X or Z can be derived, so VY is not a key.

• VX+ = {V, X}. No applicable FD produces Y, W, or Z, so VX is not a key.

• XY+ = {X, Y}. No FD applies to produce V, W, or Z, so XY is not a key.

Conclusion: VXY determines all attributes and no proper subset of VXY does, so VXY is a candidate key.