Key Concepts and Formulas

• Implication (→): $A \to B$ is false only when $A$ is true and $B$ is false. Otherwise, it is true.
• Conjunction (∧): $A \wedge B$ is true only when both $A$ and $B$ are true. Otherwise, it is false.
• Disjunction (∨): $A \vee B$ is false only when both $A$ and $B$ are false. Otherwise, it is true.
• Negation (∼): If $A$ is true, then $\sim A$ is false, and if $A$ is false, then $\sim A$ is true.

Step-by-Step Solution

Step 1: Analyze the given statement. We are given that $(p \wedge \sim q) \wedge (p \wedge r) \to \sim p \vee q$ is false. Using the truth table properties of implication, the only way an implication $A \to B$ can be false is when $A$ is true and $B$ is false. Therefore, we must have: $(p \wedge \sim q) \wedge (p \wedge r) \text{ is True, and } \sim p \vee q \text{ is False.}$

Step 2: Determine the truth values of $\sim p$ and $q$. Since $\sim p \vee q$ is false, both $\sim p$ and $q$ must be false. $\sim p = \text{False} \text{ and } q = \text{False}.$

Step 3: Determine the truth value of $p$. Since $\sim p$ is false, $p$ must be true. $p = \text{True}.$

Step 4: Determine the truth value of $\sim q$. Since $q$ is false, $\sim q$ must be true. $\sim q = \text{True}.$

Step 5: Analyze the truth values of $p \wedge \sim q$ and $p \wedge r$. Since $(p \wedge \sim q) \wedge (p \wedge r)$ is true, both $p \wedge \sim q$ and $p \wedge r$ must be true. $p \wedge \sim q = \text{True} \text{ and } p \wedge r = \text{True}.$

Step 6: Determine the truth value of $r$. We know $p = \text{True}$, and $p \wedge r = \text{True}$. Therefore, $r$ must be true. $r = \text{True}.$

Step 7: Summarize the truth values. We have found that $p = \text{True}$, $q = \text{False}$, and $r = \text{True}$.

Step 8: Check if the solution matches the given correct answer. The truth values we found are $p = T, q = F, r = T$. The given correct answer is (A) F, T, F. However, we derived T, F, T. There seems to be a mistake in the provided correct answer. Let's re-evaluate the given options. If $p=F, q=T, r=F$ (option A), then $\sim q = F$, so $p \wedge \sim q = F \wedge F = F$. Therefore $(p \wedge \sim q) \wedge (p \wedge r) = F \wedge (F \wedge F) = F$. Also, $\sim p \vee q = T \vee T = T$. So, $(p \wedge \sim q) \wedge (p \wedge r) \to \sim p \vee q$ becomes $F \to T$ which is True.

If $p=T, q=F, r=T$ (option B), then $\sim q = T$, so $p \wedge \sim q = T \wedge T = T$. Therefore $(p \wedge \sim q) \wedge (p \wedge r) = T \wedge (T \wedge T) = T$. Also, $\sim p \vee q = F \vee F = F$. So, $(p \wedge \sim q) \wedge (p \wedge r) \to \sim p \vee q$ becomes $T \to F$ which is False.

If $p=T, q=T, r=T$ (option C), then $\sim q = F$, so $p \wedge \sim q = T \wedge F = F$. Therefore $(p \wedge \sim q) \wedge (p \wedge r) = F \wedge (T \wedge T) = F$. Also, $\sim p \vee q = F \vee T = T$. So, $(p \wedge \sim q) \wedge (p \wedge r) \to \sim p \vee q$ becomes $F \to T$ which is True.

If $p=F, q=F, r=F$ (option D), then $\sim q = T$, so $p \wedge \sim q = F \wedge T = F$. Therefore $(p \wedge \sim q) \wedge (p \wedge r) = F \wedge (F \wedge F) = F$. Also, $\sim p \vee q = T \vee F = T$. So, $(p \wedge \sim q) \wedge (p \wedge r) \to \sim p \vee q$ becomes $F \to T$ which is True.

The only option that makes the given statement false is option B.

Common Mistakes & Tips

• Remember the truth tables for each logical operator (∧, ∨, ∼, →).
• The implication $A \to B$ is only false when A is true and B is false.
• Work systematically and break down the complex statement into smaller parts.

Summary

We are given that $(p \wedge \sim q) \wedge (p \wedge r) \to \sim p \vee q$ is false. This means that $(p \wedge \sim q) \wedge (p \wedge r)$ is true and $\sim p \vee q$ is false. From the second part, we deduce that $\sim p$ is false and $q$ is false, which means $p$ is true. From the first part, since $p$ is true and $p \wedge \sim q$ must be true, then $\sim q$ must be true, which is consistent with $q$ being false. Since $p$ is true and $p \wedge r$ is true, $r$ must be true. Therefore, $p=T, q=F, r=T$. This corresponds to option (B). The given correct answer (A) is incorrect.

Final Answer

The final answer is \boxed{T, F, T}, which corresponds to option (B).