Key Concepts and Formulas

• Logical Equivalence: Two statements are logically equivalent if they have the same truth value under all possible circumstances.
• Truth Tables: A truth table is a table that shows the truth value of a compound statement for all possible combinations of truth values of its component statements.
• Logical Operators: We will use the following logical operators:
  • $\wedge$: Conjunction (AND)
  • $\vee$: Disjunction (OR)
  • $\sim$: Negation (NOT)
  • $\Leftrightarrow$: Biconditional (IF AND ONLY IF)
  • $\Rightarrow$: Conditional (IF...THEN...)

Step-by-Step Solution

Step 1: State the given equivalence.

We are given that $(p \wedge r) \Leftrightarrow (p \wedge (\sim q))$ is equivalent to $(\sim p)$. This can be written as:

$(p \wedge r) \Leftrightarrow (p \wedge (\sim q)) \equiv (\sim p)$

Step 2: Use the biconditional equivalence.

The biconditional $A \Leftrightarrow B$ is equivalent to $(A \Rightarrow B) \wedge (B \Rightarrow A)$. Also, $A \Leftrightarrow B \equiv (A \wedge B) \vee (\sim A \wedge \sim B)$. In this case, we have:

$((p \wedge r) \Leftrightarrow (p \wedge (\sim q))) \equiv (\sim p)$ $((p \wedge r) \Rightarrow (p \wedge (\sim q))) \wedge ((p \wedge (\sim q)) \Rightarrow (p \wedge r)) \equiv (\sim p)$

Step 3: Analyze the implication and simplify.

We are given that the whole statement is equivalent to $\sim p$. Let's analyze what this means. If $p$ is true, then $\sim p$ is false. Thus, if $p$ is true, the statement $(p \wedge r) \Leftrightarrow (p \wedge (\sim q))$ must be false. If $p$ is false, $\sim p$ is true, so $(p \wedge r) \Leftrightarrow (p \wedge (\sim q))$ must be true.

When $p$ is false, $p \wedge r$ is false and $p \wedge (\sim q)$ is false. Therefore, $(p \wedge r) \Leftrightarrow (p \wedge (\sim q))$ is true. This aligns with $\sim p$ being true.

Step 4: Consider the case when $p$ is true.

When $p$ is true, we need $(p \wedge r) \Leftrightarrow (p \wedge (\sim q))$ to be false (because $\sim p$ is false). Since $p$ is true, $p \wedge r$ is true if and only if $r$ is true. Also, $p \wedge (\sim q)$ is true if and only if $\sim q$ is true. Therefore, we need $r \Leftrightarrow \sim q$ to be false when $p$ is true.

This means that if $r$ is true, $\sim q$ must be false (i.e., $q$ must be true), and if $r$ is false, $\sim q$ must be true (i.e., $q$ must be false). In other words, we need $r$ to always have the same truth value as $q$.

Step 5: Analyze further when $p$ is true.

Now, when $p$ is true, we require that $(p \wedge r) \Leftrightarrow (p \wedge (\sim q))$ must be false. So, when $p$ is true, $r \Leftrightarrow \sim q$ should be false. That is, $r$ and $\sim q$ must have different truth values.

If $r = p$, then when $p$ is true, $r$ is also true. Then, we need $q$ to be true (so $\sim q$ is false). In this case, we need $(p \wedge p) \Leftrightarrow (p \wedge (\sim q)) \equiv \sim p$ which simplifies to $p \Leftrightarrow (p \wedge (\sim q)) \equiv \sim p$. This becomes $p \Leftrightarrow \sim q \equiv \sim p$.

If $p$ is true, then $\sim p$ is false. So $p \Leftrightarrow \sim q$ must be false. This means that $p$ and $\sim q$ must have different truth values. So if $p$ is true, $\sim q$ is false, which means $q$ is true.

If $p$ is false, then $\sim p$ is true. So $p \Leftrightarrow \sim q$ must be true. This means $p$ and $\sim q$ must have the same truth value. Since $p$ is false, $\sim q$ must be false, so $q$ is true.

So $q$ is always true, which is a contradiction since $q$ can be either true or false.

Step 6: Re-evaluate the condition.

We need $(p \wedge r) \Leftrightarrow (p \wedge \sim q) \equiv \sim p$. Consider $p$ is True. Then $\sim p$ is False. Hence $(p \wedge r) \Leftrightarrow (p \wedge \sim q)$ must be False. $p \wedge r$ is True only when $r$ is True. $p \wedge \sim q$ is True only when $\sim q$ is True i.e., $q$ is False. So, when $p$ is True, we want $r \Leftrightarrow \sim q$ to be False. Consider $p$ is False. Then $\sim p$ is True. Hence $(p \wedge r) \Leftrightarrow (p \wedge \sim q)$ must be True. $p \wedge r$ is always False. $p \wedge \sim q$ is always False. So $(p \wedge r) \Leftrightarrow (p \wedge \sim q)$ is always True.

Thus, we only need to satisfy the condition when $p$ is True. When $p$ is True, we want $r \Leftrightarrow \sim q$ to be False. i.e., $r$ and $\sim q$ should have different truth values.

Now let's check the options: (A) $r = p$. When $p$ is True, $r$ is True. We want $\sim q$ to be False, i.e., $q$ to be True. This is consistent. (B) $r = \sim p$. When $p$ is True, $r$ is False. We want $\sim q$ to be True, i.e., $q$ to be False. This is consistent. (C) $r = q$. When $p$ is True, we want $q \Leftrightarrow \sim q$ to be False, which is always True. So $q$ can be anything. (D) $r = \sim q$. When $p$ is True, we want $\sim q \Leftrightarrow \sim q$ to be False, which is always False. So, $\sim q$ must be False.

If $r=p$, then $(p \wedge r) \Leftrightarrow (p \wedge \sim q)$ becomes $(p \wedge p) \Leftrightarrow (p \wedge \sim q)$ or $p \Leftrightarrow (p \wedge \sim q)$. Then we want $p \Leftrightarrow (p \wedge \sim q) \equiv \sim p$. This means that $p \Leftrightarrow (p \wedge \sim q)$ and $\sim p$ must have the same truth values.

If $p$ is true, we need $p \Leftrightarrow (p \wedge \sim q)$ to be false. $p \wedge \sim q$ is true only if $\sim q$ is true, i.e., $q$ is false. So if $q$ is false, $p \Leftrightarrow (p \wedge \sim q)$ is $p \Leftrightarrow p$, which is true. So we need $q$ to be true.

If $p$ is false, we need $p \Leftrightarrow (p \wedge \sim q)$ to be true. This means $p \Leftrightarrow$ false is true, which is true.

So, $r = p$ works.

Common Mistakes & Tips

• Be careful with the order of operations when evaluating logical expressions. Parentheses are very important.
• Remember the equivalences of logical operators, especially biconditional and conditional.
• When working with truth tables, be systematic to avoid errors.

Summary

We are given that $(p \wedge r) \Leftrightarrow (p \wedge (\sim q)) \equiv (\sim p)$. We analyzed the cases when $p$ is true and when $p$ is false. When $p$ is false, the equivalence is always true. When $p$ is true, we require $r \Leftrightarrow \sim q$ to be false. We then checked each of the options to see which one satisfies this condition. The correct option is $r=p$.

Final Answer

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