The logical statement (p -> ~q) <-> (p ^ q) is A) a tautology B) a contradiction C) neither a tautology nor a contradiction (D) equivalent to ~p V q

Solution

The correct option is B: a contradiction.

To determine the nature of the statement $(p \rightarrow \sim q) \leftrightarrow (p \wedge q)$, we construct a truth table to examine all value combinations of $p$ and $q$:

[ \begin{array}{|c|c|c|c|c|c|} \hline p & q & \sim q & p \rightarrow \sim q & p \wedge q & (p \rightarrow \sim q) \leftrightarrow (p \wedge q) \ \hline T & T & F & F & T & F \ T & F & T & T & F & F \ F & T & F & T & F & F \ F & F & T & T & F & F \ \hline \end{array} ]

• Column $\sim q$: Negation of $q$.
• Column $p \rightarrow \sim q$: Represents $p$ implies not $q$; true unless $p$ is true and $\sim q$ (not $q$) is false.
• Column $p \wedge q$: True if both $p$ and $q$ are true.
• The final column evaluates $(p \rightarrow \sim q) \leftrightarrow (p \wedge q)$, which checks whether the two expressions have the same truth value.

From the truth table, we observe that the final column always results in false. Therefore, the statement is a contradiction because it can never be true under any combination of truth values for $p$ and $q$. Thus, $(p \rightarrow \sim q) \leftrightarrow (p \wedge q)$ is not equivalent to $\sim p \vee q$ nor any other expression that contains a true scenario. Hence, this confirms that the correct choice is (B) a contradiction.