1. **State the problem:** Show that $P \leftrightarrow Q$ is equivalent to $\neg P \leftrightarrow \neg Q$ using both truth tables and laws of logic. 2. **Truth Table Method:** | $P$ | $Q$ | $P \leftrightarrow Q$ | $\neg P$ | $\neg Q$ | $\neg P \leftrightarrow \neg Q$ | |-----|-----|-----------------------|----------|----------|-------------------------------| | T | T | T | F | F | T | | T | F | F | F | T | F | | F | T | F | T | F | F | | F | F | T | T | T | T | From the table, the columns for $P \leftrightarrow Q$ and $\neg P \leftrightarrow \neg Q$ are identical, so they are equivalent. 3. **Laws of Logic Method:** - Recall the biconditional definition: $$P \leftrightarrow Q \equiv (P \land Q) \lor (\neg P \land \neg Q)$$ - Similarly, $$\neg P \leftrightarrow \neg Q \equiv (\neg P \land \neg Q) \lor (\neg \neg P \land \neg \neg Q)$$ - Using double negation law: $$\neg \neg P \equiv P, \quad \neg \neg Q \equiv Q$$ - Substitute back: $$\neg P \leftrightarrow \neg Q \equiv (\neg P \land \neg Q) \lor (P \land Q)$$ - By commutativity of disjunction: $$(P \land Q) \lor (\neg P \land \neg Q)$$ - This matches the expression for $P \leftrightarrow Q$, proving equivalence. **Summary:** Both truth tables and laws of logic confirm that $P \leftrightarrow Q$ is equivalent to $\neg P \leftrightarrow \neg Q$.