In context of databases consider set operations on relation R and S, and select a true statement:

Question

Accepted Answer

The correct option is (3):
The fundamental laws of set theory govern how relations, $R$ and $S$, interact in a database context. The expression $(R-S)$ represents all tuples found in relation $R$ but not in relation $S$. Conversely, the expression $(S-R)$ represents all tuples found in relation $S$ but not in relation $R$. By definition, the set $(R-S)$ contains only the elements *unique* to $R$, and the set $(S-R)$ contains only the elements *unique* to $S$. Because these two resulting sets contain elements that are mutually exclusive—that is, a tuple cannot be simultaneously in $R$ and *not* in $S$, while also being in $S$ and *not* in $R$—they can never have any elements in common. Therefore, the intersection ($\cap$) of these two disjoint sets must always result in the empty set ($\phi$), making the statement $(R-S)\cap(S-R)=\phi$ an absolute truth in set theory and relational algebra.

The incorrect option is (1):
The statement $(R-S)\cup(S-R)=\phi$ is incorrect because the union of two non-empty sets can only be the empty set if both constituent sets are themselves empty. The term $(R-S)\cup(S-R)$ is known as the Symmetric Difference, which represents all tuples unique to $R$ or unique to $S$, excluding the intersection. For this entire expression to equal the empty set ($\phi$), it would require that $R$ and $S$ be identical relations, meaning $R-S$ and $S-R$ are both $\phi$. Since we must consider the general case where $R$ and $S$ are often different relations with unique tuples, this equality does not hold true universally.

The incorrect option is (2):
The statement $(R-S)\cup(S-R)=R\cap S$ is logically flawed as it equates the symmetric difference to the intersection of $R$ and $S$. The intersection $R\cap S$ includes all tuples that are *common* to both relations. In stark contrast, the symmetric difference $(R-S)\cup(S-R)$ explicitly contains only the tuples that are *not* common to both, as it excludes the intersection entirely. Therefore, one set containing only common elements can never equal the set containing only non-common elements, except in the trivial and uninteresting case where both $R$ and $S$ are empty.

The incorrect option is (4):
The statement $(R-S)\cup(S-R)=R\cup S$ is generally false because the expression on the left, the Symmetric Difference, deliberately excludes all elements present in the intersection $R\cap S$. On the other hand, the expression on the right, $R\cup S$, represents the full combination of all tuples in both $R$ and $S$ and therefore *includes* the intersection. The resulting set from the symmetric difference is a proper subset of the full union, unless the relations $R$ and $S$ are completely disjoint. Since the two sides are only equal in a special case, the statement is not universally true.