Key Concepts and Formulas

• Logical AND ($\wedge$): $p \wedge q$ is true only if both $p$ and $q$ are true. Otherwise, it's false.
• Logical Equivalence ($\leftrightarrow$): $p \leftrightarrow q$ is true if both $p$ and $q$ have the same truth value (both true or both false). Otherwise, it's false.
• Tautology: A statement that is always true, regardless of the truth values of its components.
• Logical OR ($\vee$): $p \vee q$ is true if at least one of $p$ or $q$ is true. It's false only if both are false.
• Logical Implication ($\to$): $p \to q$ is true if $p$ is false, or if $q$ is true (or both). It is only false when $p$ is true and $q$ is false.

Step-by-Step Solution

Step 1: Analyze the given information. We are given that $q$ is false (F) and $(p \wedge q) \leftrightarrow r$ is true (T).

Step 2: Determine the truth value of $p \wedge q$. Since $q$ is false, $p \wedge q$ must also be false, regardless of the truth value of $p$. This is because for $p \wedge q$ to be true, both $p$ and $q$ must be true, and we know $q$ is false. So, $p \wedge q = F$.

Step 3: Determine the truth value of $r$. We know that $(p \wedge q) \leftrightarrow r$ is true, and we just found that $p \wedge q$ is false. Therefore, $r$ must also be false for the equivalence to be true. If $r$ were true, then $(p \wedge q) \leftrightarrow r$ would be false. So, $r = F$.

Step 4: Evaluate the truth values of the given options. Now that we know $q = F$ and $r = F$, we can analyze each option to see which one is a tautology.

(A) $p \wedge r$ Since $r$ is false, $p \wedge r$ is always false, regardless of the truth value of $p$. Thus, this is not a tautology.

(B) $(p \vee r) \to (p \wedge r)$ Since $r$ is false, this becomes $(p \vee F) \to (p \wedge F)$, which simplifies to $p \to F$. This statement is false when $p$ is true and false when $p$ is false. Thus, it is not a tautology.

(C) $p \vee r$ Since $r$ is false, this becomes $p \vee F$, which simplifies to $p$. This statement is true when $p$ is true and false when $p$ is false. Thus, it is not a tautology.

(D) $(p \wedge r) \to (p \vee r)$ Since $r$ is false, $p \wedge r$ is always false. Therefore, the statement becomes $F \to (p \vee F)$, which is equivalent to $F \to p$. Since $F \to x$ is always true, this statement is a tautology.

Step 5: Verify the correct answer. We need to check if the answer we arrived at matches the given correct answer, which is option (D).

Common Mistakes & Tips

• Remember the truth tables for the logical operators. A quick reference can prevent errors.
• When dealing with equivalences, remember that both sides must have the same truth value for the entire statement to be true.
• A statement is a tautology if and only if it is always true, regardless of the truth values of its components. Don't settle for just one case where the statement is true.

Summary

We are given that $q$ is false and $(p \wedge q) \leftrightarrow r$ is true. From this, we deduced that $r$ is also false. Then, we evaluated each of the given options with $r = F$ to determine which one is a tautology. We found that option (D), $(p \wedge r) \to (p \vee r)$, is a tautology because when $r$ is false, $p \wedge r$ is always false, making the implication always true.

Final Answer

The final answer is \boxed{(p \wedge r) \to (p \vee r)}, which corresponds to option (D).