Consider two relations R(A,B) and S(E,C). A is primary key and E is a FK referring A. Which of the following operations never violate FK constraint?

The correct options are Insert in R and Delete from S.

To understand why these operations never violate the Foreign Key (FK) constraint, let us analyze the relationship between the two relations, $R(A,B)$ and $S(E,C)$.

Here, $A$ is the Primary Key (PK) of relation $R$, and $E$ is a Foreign Key in relation $S$ that references $A$ in relation $R$. The Foreign Key constraint requires that every value of $E$ in $S$ must either be null (if allowed) or exist as a value of $A$ in $R$. That is, ${\pi }_E\left(S\right)\subseteq {\pi }_A\left(R\right)$ (excluding null values).

Let us evaluate each operation to see how it affects this constraint:

1. Insert in R:
When we insert a new tuple into the referenced relation $R$, we add a new primary key value to $A$. This expands the set of valid values that the foreign key $E$ can reference. Since it does not remove any existing primary keys, it cannot cause any foreign key in $S$ to point to a non-existent primary key in $R$. Therefore, inserting into $R$ can never violate the foreign key constraint.

2. Delete from S:
When we delete a tuple from the referencing relation $S$, we remove a foreign key value $E$. Since the constraint only requires that existing foreign key values in $S$ must exist in $R$, removing a row from $S$ simply reduces the referencing set. It cannot create an orphaned reference because we are removing references, not adding them. Therefore, deleting from $S$ can never violate the foreign key constraint.

For completeness, let us look at the other options:
- Insert in S: If we insert a tuple into $S$ with a foreign key value $E$ that does not exist in $R$, the constraint is violated.
- Delete from R: If we delete a tuple from $R$ whose primary key $A$ is currently being referenced by some tuple's foreign key $E$ in $S$, the constraint is violated (unless cascade actions are specified).