Key Concepts and Formulas

• Implication: $p \to q$ is FALSE only when $p$ is TRUE and $q$ is FALSE. Otherwise, it's TRUE.
• Negation: $\sim p$ is TRUE if $p$ is FALSE, and FALSE if $p$ is TRUE.
• Conjunction: $p \wedge q$ is TRUE only when both $p$ and $q$ are TRUE. Otherwise, it's FALSE.
• Disjunction: $p \vee q$ is FALSE only when both $p$ and $q$ are FALSE. Otherwise, it's TRUE.

Step-by-Step Solution

Step 1: Analyze the given condition

We are given that $p \to (\sim p \vee q)$ is FALSE. Using the truth table for implication, this means that $p$ must be TRUE and $(\sim p \vee q)$ must be FALSE. $p \to (\sim p \vee q) = F$ This implies: $p = T \quad \text{and} \quad (\sim p \vee q) = F$

Step 2: Determine the truth value of p and q

Since $p = T$, we have $\sim p = F$. Substituting this into $(\sim p \vee q) = F$, we get: $(F \vee q) = F$ For the disjunction to be FALSE, both terms must be FALSE. Therefore, $q = F$. $q = F$ So, we have $p = T$ and $q = F$.

Step 3: Evaluate proposition P1

$P1: \sim (p \to \sim q)$ Substitute $p = T$ and $q = F$: $P1: \sim (T \to \sim F)$ Since $\sim F = T$, we have: $P1: \sim (T \to T)$ Since $T \to T = T$, we have: $P1: \sim (T) = F$ Thus, P1 is FALSE.

Step 4: Evaluate proposition P2

$P2: (p \wedge \sim q) \wedge ((\sim p) \vee q)$ Substitute $p = T$ and $q = F$: $P2: (T \wedge \sim F) \wedge ((\sim T) \vee F)$ Since $\sim F = T$ and $\sim T = F$, we have: $P2: (T \wedge T) \wedge (F \vee F)$ $P2: (T) \wedge (F)$ $P2: F$ Thus, P2 is FALSE.

Step 5: Determine which option is correct

We found that P1 is FALSE and P2 is FALSE.

Common Mistakes & Tips

• Remember that $p \to q$ is only FALSE when $p$ is TRUE and $q$ is FALSE. It's TRUE in all other cases.
• Pay close attention to the order of operations and the meaning of each logical connective ($\sim, \wedge, \vee, \to$).
• When evaluating compound propositions, substitute the truth values of the variables and simplify step by step.

Summary

We were given that $p \to (\sim p \vee q)$ is FALSE, which implied that $p$ is TRUE and $\sim p \vee q$ is FALSE. This allowed us to deduce that $p = T$ and $q = F$. Substituting these values into the expressions for P1 and P2, we found that both P1 and P2 are FALSE. Therefore, the correct option is that both P1 and P2 are FALSE.

Final Answer

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