1. **State the problem:** Prove the logical implication $ (p \lor q) \to (p \land q) $ using laws of logic. 2. **Recall the implication equivalence:** An implication $A \to B$ is logically equivalent to $\neg A \lor B$. 3. **Rewrite the implication:** $$ (p \lor q) \to (p \land q) \equiv \neg (p \lor q) \lor (p \land q) $$ 4. **Apply De Morgan's law to $\neg (p \lor q)$:** $$ \neg (p \lor q) \equiv \neg p \land \neg q $$ 5. **Substitute back:** $$ (\neg p \land \neg q) \lor (p \land q) $$ 6. **Analyze the expression:** This is a disjunction of two conjunctions: one where both $p$ and $q$ are false, and one where both are true. 7. **Check if the expression is a tautology:** For the original implication to be true, this expression must be true for all truth values of $p$ and $q$. 8. **Test truth values:** - If $p = \text{true}$ and $q = \text{false}$, then $\neg p \land \neg q = \text{false}$ and $p \land q = \text{false}$, so the whole expression is false. 9. **Conclusion:** Since the expression is not always true, the implication $ (p \lor q) \to (p \land q) $ is **not a tautology** and thus cannot be proven true by laws of logic. **Final answer:** The implication $ (p \lor q) \to (p \land q) $ is false in general and cannot be proven true by logical laws.