The following functional dependencies hold true for the relational schema {V,…

Goal: find an irreducible (minimal) equivalent set of functional dependencies for the given schema.

1. Step 1: Split right-hand sides so each FD has a single attribute on the right.

   From Y → VX produce Y → V and Y → X. Now the set is: V → W, VW → X, Y → V, Y → X, Y → Z.

2. Step 2: Minimize left-hand sides (remove extraneous attributes).

   Test VW → X: check if X is in W+ (using all FDs). W+ = {W}, so X is not there. Check if X is in V+; V+ = {V, W} using V → W, and then {V, W, X} via VW → X, so W is extraneous in VW → X. Therefore reduce VW → X to V → X.

3. Step 3: Remove redundant dependencies.

   After reduction we have V → W, V → X, Y → V, Y → X, Y → Z. Note that Y → X is implied by Y → V together with V → X (transitivity), so Y → X is redundant and can be removed.

4. Step 4: Verify minimality.

   The remaining set is V → W, V → X, Y → V, Y → Z. Each FD has a single attribute on the right, no left-hand attribute is extraneous, and none of the FDs can be derived from the others, so the set is minimal (irreducible).

Final irreducible (minimal) equivalent:

• V → W

• V → X

• Y → V

• Y → Z

This set is irreducible because every dependency has a single attribute on the right, no left-hand attribute is extraneous, and there are no redundant dependencies.