Biconditional 2E7678

1. **State the problem:** We want to understand and verify the biconditional statement $p \leftrightarrow q$, which means "$p$ if and only if $q$". 2. **Definition:** The biconditional $p \leftrightarrow q$ is defined as the conjunction of two conditionals: $$p \leftrightarrow q \equiv (p \to q) \wedge (q \to p)$$ This means $p$ implies $q$ and $q$ implies $p$. 3. **Truth table:** To verify this, we construct the truth table for $p$, $q$, and $p \leftrightarrow q$: | $p$ | $q$ | $p \to q$ | $q \to p$ | $(p \to q) \wedge (q \to p)$ | |-----|-----|-----------|-----------|-------------------------------| | T | T | T | T | T | | T | F | F | T | F | | F | T | T | F | F | | F | F | T | T | T | 4. **Explanation:** - When both $p$ and $q$ are true, both implications are true, so the biconditional is true. - When $p$ is true and $q$ is false, $p \to q$ is false, so the biconditional is false. - When $p$ is false and $q$ is true, $q \to p$ is false, so the biconditional is false. - When both $p$ and $q$ are false, both implications are true, so the biconditional is true. 5. **Conclusion:** The biconditional $p \leftrightarrow q$ is true exactly when $p$ and $q$ have the same truth value. **Final answer:** The biconditional $p \leftrightarrow q$ is logically equivalent to $(p \to q) \wedge (q \to p)$ and is true when $p$ and $q$ are both true or both false.