GATE CS 2015 Set 1 Q.24 (CS Q.14): Propositional and First-Order Logic

Which one of the following is NOT equivalent to $p \leftrightarrow q$ ?

$(\neg p \lor q) \land (p \lor \neg q)$

$(\neg p \lor q) \land (q \to p)$

$(\neg p \land q) \lor (p \land \neg q)$

$(\neg p \land \neg q) \lor (p \land q)$

Official answer and solution

Solution

Compare the truth table and algebraic forms of each candidate proposition with the biconditional $p \leftrightarrow q$ .

The biconditional $p \leftrightarrow q$ is true when $p$ and $q$ have identical truth values ( $T, T$ or $F, F$ ), and false when they differ.

Analyzing each option:

Option A: $(\neg p \lor q) \land (p \lor \neg q) \equiv (p \to q) \land (q \to p) \equiv p \leftrightarrow q$ . (Equivalent)
Option B: $(\neg p \lor q) \land (q \to p) \equiv (p \to q) \land (q \to p) \equiv p \leftrightarrow q$ . (Equivalent)
Option D: $(\neg p \land \neg q) \lor (p \land q)$ . This is the disjunctive normal form of $p \leftrightarrow q$ . (Equivalent)
Option C: $(\neg p \land q) \lor (p \land \neg q)$ . This proposition is true precisely when $p$ and $q$ have opposite truth values. It represents the exclusive-OR (XOR) operation $p \oplus q \equiv \neg(p \leftrightarrow q)$ .

Therefore, option C is NOT equivalent to $p \leftrightarrow q$ .

Final answer

Common trap

Confusing the biconditional (XNOR) with the exclusive-OR (XOR) formula in option C.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

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