Problem: Verify that $ (p \vee q) \wedge (p \vee r) $ is logically equivalent to $ p \vee (q \wedge r) $. 1. Approach: We prove the equivalence by showing each direction separately, and we will also explain how the given truth table matches the reasoning. 2. Proof of $ (p \vee q) \wedge (p \vee r) \implies p \vee (q \wedge r) $. Assume $ (p \vee q) \wedge (p \vee r) $ is true. Either $p$ is true or $p$ is false. If $p$ is true then $p \vee (q \wedge r)$ is true. If $p$ is false then both $p \vee q$ and $p \vee r$ must be true, so $q$ is true and $r$ is true, hence $q \wedge r$ is true, and thus $p \vee (q \wedge r)$ is true. Therefore in all cases $ (p \vee q) \wedge (p \vee r) \implies p \vee (q \wedge r)$. 3. Proof of $ p \vee (q \wedge r) \implies (p \vee q) \wedge (p \vee r) $. Assume $p \vee (q \wedge r)$ is true. If $p$ is true then both $p \vee q$ and $p \vee r$ are true, so their conjunction is true. If $q \wedge r$ is true then $q$ is true and $r$ is true, so again both $p \vee q$ and $p \vee r$ are true, and their conjunction is true. Therefore $p \vee (q \wedge r) \implies (p \vee q) \wedge (p \vee r)$. 4. Conclusion: Since each formula implies the other we have the biconditional $ (p \vee q) \wedge (p \vee r) \iff p \vee (q \wedge r) $. This means the equivalence is a logical truth (the biconditional is true for every assignment of truth values to $p,q,r$). 5. Relation to the truth table: The rows of the provided truth table assign every combination of truth values to $p,q,r$ and compute both expressions. For each row the truth values of $ (p \vee q) \wedge (p \vee r) $ and $ p \vee (q \wedge r) $ coincide, which confirms the equivalence by exhaustive checking. Final answer: $ (p \vee q) \wedge (p \vee r) $ is logically equivalent to $ p \vee (q \wedge r) $.