GATE CS 1994 Q3.13 — Propositional and First Order Logic

To determine whether $p \Longleftrightarrow q$ does not imply $p \to \lnot q$, we can construct a truth table for both propositions.

Let:

• Premise: $A = (p \Longleftrightarrow q)$
• Conclusion: $B = (p \to \lnot q)$

An implication $A \implies B$ holds if and only if there is no row in the truth table where $A$ is True and $B$ is False.

Analysis:

• In the first row, when both $p$ and $q$ are True, the premise $p \Longleftrightarrow q$ is True, but the conclusion $p \to \lnot q$ is False.
• Since we found a counterexample where the premise is True and the conclusion is False, the implication does not hold.

Therefore, the statement "$p \Longleftrightarrow q$ does not imply $p \to \lnot q$" is True (represented as 1).