Key Concepts and Formulas

• Negation of an implication: $\sim(p \Rightarrow q) \equiv p \wedge \sim q$
• De Morgan's Laws: $\sim(p \vee q) \equiv \sim p \wedge \sim q$ and $\sim(p \wedge q) \equiv \sim p \vee \sim q$
• Distributive Laws: $p \wedge (q \vee r) \equiv (p \wedge q) \vee (p \wedge r)$ and $p \vee (q \wedge r) \equiv (p \vee q) \wedge (p \vee r)$

Step-by-Step Solution

Step 1: State the given statement and express its negation. We are given the statement $(p \vee r) \Rightarrow (q \vee r)$. We want to find its negation, which is $\sim((p \vee r) \Rightarrow (q \vee r))$.

Step 2: Apply the negation of implication formula. Using the formula $\sim(p \Rightarrow q) \equiv p \wedge \sim q$, we have: $\sim((p \vee r) \Rightarrow (q \vee r)) \equiv (p \vee r) \wedge \sim(q \vee r)$

Step 3: Apply De Morgan's Law to simplify $\sim(q \vee r)$. Using De Morgan's Law, $\sim(q \vee r) \equiv \sim q \wedge \sim r$. Substituting this into the expression from Step 2, we get: $(p \vee r) \wedge (\sim q \wedge \sim r)$

Step 4: Apply the associative law. We can rearrange the expression as follows: $(p \vee r) \wedge (\sim q \wedge \sim r) \equiv (p \vee r) \wedge (\sim r \wedge \sim q)$

Step 5: Apply the distributive law to expand $(p \vee r) \wedge (\sim r \wedge \sim q)$. Since $\wedge$ is associative, we can write $(p \vee r) \wedge (\sim q \wedge \sim r)$ as $(p \vee r) \wedge (\sim r \wedge \sim q)$. Applying the distributive law, $a \wedge (b \wedge c) = (a \wedge b) \wedge c$, we can associate $(\sim q \wedge \sim r)$ as a single term. Now, using $p \wedge (q \vee r) \equiv (p \wedge q) \vee (p \wedge r)$, we can consider $(\sim r \wedge \sim q)$ as a single term. Then, $(p \vee r) \wedge (\sim q \wedge \sim r) \equiv (p \vee r) \wedge (\sim r \wedge \sim q)$. We now rearrange the terms using the commutative property of $\wedge$: $(p \vee r) \wedge (\sim q \wedge \sim r) \equiv (p \vee r) \wedge (\sim r \wedge \sim q) \equiv (p \vee r) \wedge (\sim r \wedge \sim q)$. We want to obtain $p \wedge \sim q \wedge \sim r$. So let's rewrite as $(p \vee r) \wedge (\sim r \wedge \sim q) = (p \vee r) \wedge (\sim q \wedge \sim r)$. We can rewrite this as $(p \vee r) \wedge (\sim r \wedge \sim q) \equiv (p \vee r) \wedge (\sim r) \wedge (\sim q)$. Now distribute the $\sim r$ across the $(p \vee r)$ term. $(p \vee r) \wedge \sim r \equiv (p \wedge \sim r) \vee (r \wedge \sim r) \equiv (p \wedge \sim r) \vee F \equiv (p \wedge \sim r)$. Therefore, $(p \vee r) \wedge (\sim r) \wedge (\sim q) \equiv (p \wedge \sim r) \wedge (\sim q)$.

Step 6: Simplify the expression. So we have $(p \wedge \sim r) \wedge (\sim q)$. Since $\wedge$ is associative and commutative, we can write this as: $p \wedge \sim r \wedge \sim q \equiv p \wedge \sim q \wedge \sim r$

Common Mistakes & Tips

• Remember the correct negation of an implication. It's a common mistake to negate both parts of the implication.
• Be careful when applying De Morgan's Laws. Ensure you negate both terms and change the operator.
• When dealing with multiple logical operators, use parentheses to maintain clarity and avoid ambiguity.

Summary

We started with the negation of the given statement $(p \vee r) \Rightarrow (q \vee r)$. We applied the negation of implication formula and De Morgan's Law to simplify the expression. We then used the distributive and associative laws to arrive at the final simplified expression $p \wedge \sim q \wedge \sim r$. This corresponds to option (A).

Final Answer The final answer is \boxed{p \wedge \sim q \wedge \sim r}, which corresponds to option (A).