1. The problem is to prove that the negation of $pvq$ is equivalent to the negation of $p \wedge q$, specifically: $$\neg(p \lor q) \equiv \neg (p \wedge q)$$ 2. First, recall the definition of the operators: - $p \lor q$ means $p$ OR $q$. - $p \wedge q$ means $p$ AND $q$. - $\neg$ means NOT. 3. We want to show: $$\neg(p \lor q) \equiv \neg p \wedge \neg q$$ and $$\neg (p \wedge q) \equiv \neg p \lor \neg q$$ 4. Using De Morgan's laws: - The negation of a disjunction is the conjunction of the negations: $$\neg(p \lor q) = \neg p \wedge \neg q$$ - The negation of a conjunction is the disjunction of the negations: $$\neg(p \wedge q) = \neg p \lor \neg q$$ 5. So the original statement $\neg p v q \equiv \neg (p \wedge q)$ is incorrect as stated, unless interpreted carefully. 6. Correctly, applying De Morgan's laws, $$\neg (p \lor q) \equiv \neg p \wedge \neg q$$ $$\neg (p \wedge q) \equiv \neg p \lor \neg q$$ 7. Therefore, $\neg (p \vee q)$ is NOT equivalent to $\neg (p \wedge q)$. Final answer: $$\neg(p \lor q) \equiv \neg p \wedge \neg q \quad \text{and} \quad \neg (p \wedge q) \equiv \neg p \lor \neg q$$