The following functional dependencies hold true for the relational schema $R\{V, W, X, Y, Z\}$: $$\begin{aligned} V &\to W VW &\to X Y &\to VX Y &\to Z \end{aligned}$$ Which of the following is irreducible equivalent for this set of functional dependencies?

To find the irreducible equivalent (canonical / minimal cover) of the given set of functional dependencies $F$:

1. Decompose the right-hand side (RHS) to single attributes:

   • $V \to W$

   • $VW \to X$

   • $Y \to V$

   • $Y \to X$

   • $Y \to Z$

2. Check for extraneous attributes on the left-hand side (LHS):

   • In $VW \to X$, test if $W$ is extraneous: Compute the attribute closure of $V$ under the remaining dependencies:

     $$V^+ = \{V, W\}$$

     Since $W \in V^+$, attribute $W$ is extraneous in $VW \to X$. Replace $VW \to X$ with $V \to X$.

   The FD set becomes:

   $$\{V \to W,\; V \to X,\; Y \to V,\; Y \to X,\; Y \to Z\}$$

3. Remove redundant functional dependencies:

   • Test $Y \to X$: Compute $Y^+$ using the other dependencies $\{V \to W, V \to X, Y \to V, Y \to Z\}$:

     $$Y^+ = \{Y\} \xrightarrow{Y \to V} \{Y, V\} \xrightarrow{V \to X} \{Y, V, X\} \dots$$

     Since $X \in Y^+$, $Y \to X$ is redundant and can be eliminated.

   • Testing the remaining FDs shows that none of $V \to W$, $V \to X$, $Y \to V$, or $Y \to Z$ is redundant.

Hence, the minimal/irreducible cover is:

This matches option A.