Key Concepts and Formulas

• Tautology: A statement that is always true, regardless of the truth values of its components.
• Implication (→): $p \to q$ is equivalent to $\sim p \vee q$.
• De Morgan's Laws:
  • $\sim (p \wedge q) \equiv \sim p \vee \sim q$
  • $\sim (p \vee q) \equiv \sim p \wedge \sim q$
• Double Negation: $\sim (\sim p) \equiv p$

Step-by-Step Solution

We need to determine which of the given options is a tautology. We will analyze each option.

Option (A): ~(p $\wedge$ ~q) $\to$ p $\vee$ q

Step 1: Rewrite the implication using the equivalence $p \to q \equiv \sim p \vee q$. We have $\sim (p \wedge \sim q) \to (p \vee q) \equiv \sim (\sim (p \wedge \sim q)) \vee (p \vee q)$.

Step 2: Simplify the double negation. $\sim (\sim (p \wedge \sim q)) \vee (p \vee q) \equiv (p \wedge \sim q) \vee (p \vee q)$.

Step 3: Distribute $p$ over the disjunction. $(p \wedge \sim q) \vee (p \vee q) \equiv [(p \wedge \sim q) \vee p] \vee q$.

Step 4: Use the distributive law $a \vee (b \wedge c) = (a \vee b) \wedge (a \vee c)$. Thus, $p \vee (p \wedge \sim q) = p$. $[(p \wedge \sim q) \vee p] \vee q \equiv [p \vee (p \wedge \sim q)] \vee q \equiv p \vee q$.

Step 5: Simplify the expression. This step is not needed as the expression is already simpler. However, note that $p \vee (p \wedge \sim q) = p$ because $p$ is true regardless of $\sim q$.

Step 6: Since $p \vee q$ is not a tautology, we need to re-examine the initial transformation. $\sim (p \wedge \sim q) \to (p \vee q) \equiv \sim (\sim (p \wedge \sim q)) \vee (p \vee q) \equiv (p \wedge \sim q) \vee (p \vee q)$. We can rewrite $(p \wedge \sim q) \vee (p \vee q)$ as $(p \wedge \sim q) \vee p \vee q$.

Step 7: Use the absorption law $p \vee (p \wedge q) = p$. In our case, $p \vee (p \wedge \sim q) = p$. So we have $p \vee (p \wedge \sim q) \vee q \equiv p \vee q$. This is still not a tautology. We need to revisit our steps.

Step 8: Let's re-examine the original statement and apply De Morgan's Law: $\sim (p \wedge \sim q) \to (p \vee q) \equiv (\sim p \vee \sim (\sim q)) \to (p \vee q) \equiv (\sim p \vee q) \to (p \vee q)$.

Step 9: Convert the implication to a disjunction: $(\sim p \vee q) \to (p \vee q) \equiv \sim (\sim p \vee q) \vee (p \vee q)$.

Step 10: Apply De Morgan's Law: $\sim (\sim p \vee q) \vee (p \vee q) \equiv (\sim (\sim p) \wedge \sim q) \vee (p \vee q) \equiv (p \wedge \sim q) \vee (p \vee q)$.

Step 11: Apply the distributive property: $(p \wedge \sim q) \vee (p \vee q) \equiv [(p \wedge \sim q) \vee p] \vee q \equiv [p \vee (p \wedge \sim q)] \vee q$.

Step 12: Apply the absorption law: $p \vee (p \wedge \sim q) = p$. $[p \vee (p \wedge \sim q)] \vee q \equiv p \vee q$.

Step 13: The expression $p \vee q$ is NOT a tautology. So, Option A is NOT a tautology.

Option (B): ~(p $\vee$ ~q) $\to$ p $\vee$ q

Step 1: Rewrite the implication using the equivalence $p \to q \equiv \sim p \vee q$. We have $\sim (p \vee \sim q) \to (p \vee q) \equiv \sim (\sim (p \vee \sim q)) \vee (p \vee q)$.

Step 2: Simplify the double negation. $\sim (\sim (p \vee \sim q)) \vee (p \vee q) \equiv (p \vee \sim q) \vee (p \vee q)$.

Step 3: Simplify the expression. $(p \vee \sim q) \vee (p \vee q) \equiv p \vee \sim q \vee p \vee q \equiv p \vee p \vee q \vee \sim q \equiv p \vee (q \vee \sim q)$.

Step 4: Since $q \vee \sim q$ is always true, we have $q \vee \sim q \equiv T$. $p \vee (q \vee \sim q) \equiv p \vee T \equiv T$. Thus, Option B is a tautology.

Option (C): ~(p $\vee$ ~q) $\to$ p $\wedge$ q

Step 1: Rewrite the implication using the equivalence $p \to q \equiv \sim p \vee q$. We have $\sim (p \vee \sim q) \to (p \wedge q) \equiv \sim (\sim (p \vee \sim q)) \vee (p \wedge q)$.

Step 2: Simplify the double negation. $\sim (\sim (p \vee \sim q)) \vee (p \wedge q) \equiv (p \vee \sim q) \vee (p \wedge q)$.

Step 3: Apply the distributive property. $(p \vee \sim q) \vee (p \wedge q) \equiv p \vee \sim q \vee (p \wedge q) \equiv p \vee (\sim q \vee (p \wedge q))$.

Step 4: Apply the distributive property again. $p \vee (\sim q \vee (p \wedge q)) \equiv p \vee ((\sim q \vee p) \wedge (\sim q \vee q)) \equiv p \vee ((\sim q \vee p) \wedge T) \equiv p \vee (\sim q \vee p) \equiv p \vee p \vee \sim q \equiv p \vee \sim q$. Since $p \vee \sim q$ is not a tautology, Option C is not a tautology.

Option (D): p $\vee$ (~q) $\to$ p $\wedge$ q

Step 1: Rewrite the implication using the equivalence $p \to q \equiv \sim p \vee q$. We have $(p \vee \sim q) \to (p \wedge q) \equiv \sim (p \vee \sim q) \vee (p \wedge q)$.

Step 2: Apply De Morgan's Law. $\sim (p \vee \sim q) \vee (p \wedge q) \equiv (\sim p \wedge \sim (\sim q)) \vee (p \wedge q) \equiv (\sim p \wedge q) \vee (p \wedge q)$.

Step 3: Apply the distributive property. $(\sim p \wedge q) \vee (p \wedge q) \equiv (\sim p \vee p) \wedge q \equiv T \wedge q \equiv q$. Since $q$ is not a tautology, Option D is not a tautology.

It appears there was an error in the provided correct answer. Option B is the tautology.

Common Mistakes & Tips

• Carefully apply De Morgan's Laws and the implication equivalence.
• Use truth tables to check your work if necessary.
• Remember the absorption laws: $p \vee (p \wedge q) \equiv p$ and $p \wedge (p \vee q) \equiv p$.

Summary

We analyzed each of the given options to determine which one is a tautology. By applying logical equivalences and simplifications, we found that option (B), $\sim (p \vee \sim q) \to (p \vee q)$, simplifies to $T$, which is a tautology.

Final Answer

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