Superkeys in $R(V,W,X,Y,Z)$ Given Primary Key $VY$

We are given a relation $R(V,W,X,Y,Z)$ with primary key $VY$. A superkey is any attribute set whose closure contains all attributes of $R$. Since $VY$ is a primary key, $VY \to V,W,X,Y,Z$, and therefore any superset of $VY$ is also a superkey (by the property of attribute closure / functional dependency).2

For each option, we check whether it is guaranteed to functionally determine all attributes (i.e., whether its closure contains $V,W,X,Y,Z$). The quickest reasoning here is:

• If an option contains both $V$ and $Y$, then it is a superset of the primary key $VY$, hence a superkey.
• If it misses either $V$ or $Y$, it cannot be concluded to be a superkey from the fact that $VY$ is a key (and in standard exam settings, that omission means it is not a superkey).

Options:

1. (i) $VXYZ$ contains $V$ and $Y$ ⇒ superkey
2. (ii) $VWXZ$ misses $Y$ ⇒ not a superkey
3. (iii) $VWXY$ contains $V$ and $Y$ ⇒ superkey
4. (iv) $VWXYZ$ contains $V$ and $Y$ ⇒ superkey

Therefore, the option that is not a superkey is (ii) $VWXZ$.

Key terms

• superkey
• candidate key
• attribute closure
• functional dependency

How to verify systematically (attribute closure method)

Even though the “superset of the primary key” shortcut is sufficient here, the standard rigorous method is to compute attribute closure for each option and test whether it includes all attributes $R$.

The algorithm:

• For a set $S$, compute $S^+$ using the given functional dependencies (FDs).
• $S$ is a superkey iff $S^+ = \{V,W,X,Y,Z\}$ (or at least contains all those attributes).

This is grounded in the theorem: if $X$ is a candidate key, then $X$’s supersets are superkeys; and closure-based superkey testing is the formal definition.2