Key Concepts and Formulas

• Implication: $P \Rightarrow Q \equiv \sim P \vee Q$
• De Morgan's Laws: $\sim (P \wedge Q) \equiv \sim P \vee \sim Q$ and $\sim (P \vee Q) \equiv \sim P \wedge \sim Q$
• Tautology: A statement that is always true, regardless of the truth values of its components.

Step-by-Step Solution

Step 1: Simplify the left-hand side (LHS) of all the options. We are given the LHS as $(P \Rightarrow Q) \wedge \sim Q$. We will use the implication equivalence to rewrite the implication and then use distribution. $(P \Rightarrow Q) \wedge \sim Q \equiv (\sim P \vee Q) \wedge \sim Q$

Step 2: Apply the distributive law. We distribute $\sim Q$ over the disjunction: $(\sim P \vee Q) \wedge \sim Q \equiv (\sim P \wedge \sim Q) \vee (Q \wedge \sim Q)$

Step 3: Simplify using the contradiction $Q \wedge \sim Q \equiv F$ (False). $(\sim P \wedge \sim Q) \vee (Q \wedge \sim Q) \equiv (\sim P \wedge \sim Q) \vee F \equiv \sim P \wedge \sim Q$ Therefore, the LHS of all options simplifies to $\sim P \wedge \sim Q$.

Step 4: Analyze option (A): $((P \Rightarrow Q) \wedge \sim Q) \Rightarrow (P \wedge Q)$. We want to determine if $(\sim P \wedge \sim Q) \Rightarrow (P \wedge Q)$ is a tautology. Using the implication equivalence: $(\sim P \wedge \sim Q) \Rightarrow (P \wedge Q) \equiv \sim (\sim P \wedge \sim Q) \vee (P \wedge Q)$ Applying De Morgan's law: $\sim (\sim P \wedge \sim Q) \vee (P \wedge Q) \equiv (P \vee Q) \vee (P \wedge Q)$ Using the absorption law, $P \vee (P \wedge Q) \equiv P$. Therefore, $(P \vee Q) \vee (P \wedge Q) \equiv P \vee Q$

Step 5: Check if $P \vee Q$ is a tautology. $P \vee Q$ is not a tautology. If both P and Q are false, then $P \vee Q$ is false. So option (A) is incorrect in the original solution.

Step 6: Analyze option (B): $((P \Rightarrow Q) \wedge \sim Q) \Rightarrow Q$. We want to determine if $(\sim P \wedge \sim Q) \Rightarrow Q$ is a tautology. $(\sim P \wedge \sim Q) \Rightarrow Q \equiv \sim(\sim P \wedge \sim Q) \vee Q$ Applying De Morgan's law: $\sim(\sim P \wedge \sim Q) \vee Q \equiv (P \vee Q) \vee Q \equiv P \vee Q$ $P \vee Q$ is not a tautology, so option (B) is incorrect in the original solution.

Step 7: Analyze option (C): $((P \Rightarrow Q) \wedge \sim Q) \Rightarrow P$. We want to determine if $(\sim P \wedge \sim Q) \Rightarrow P$ is a tautology. $(\sim P \wedge \sim Q) \Rightarrow P \equiv \sim(\sim P \wedge \sim Q) \vee P$ Applying De Morgan's law: $\sim(\sim P \wedge \sim Q) \vee P \equiv (P \vee Q) \vee P \equiv P \vee Q$ $P \vee Q$ is not a tautology, so option (C) is incorrect in the original solution.

Step 8: Analyze option (D): $((P \Rightarrow Q) \wedge \sim Q) \Rightarrow \sim P$. We want to determine if $(\sim P \wedge \sim Q) \Rightarrow \sim P$ is a tautology. $(\sim P \wedge \sim Q) \Rightarrow \sim P \equiv \sim(\sim P \wedge \sim Q) \vee \sim P$ Applying De Morgan's law: $\sim(\sim P \wedge \sim Q) \vee \sim P \equiv (P \vee Q) \vee \sim P \equiv (P \vee \sim P) \vee Q \equiv T \vee Q \equiv T$ Since the expression simplifies to T (True), it is a tautology.

Common Mistakes & Tips

• Carefully apply De Morgan's laws and the implication equivalence. A small error in applying these rules can lead to an incorrect answer.
• Remember that $P \vee \sim P$ is always true (Tautology) and $P \wedge \sim P$ is always false (Contradiction).
• When simplifying, look for opportunities to apply the absorption law.

Summary

We simplified the left-hand side of all the given options to $\sim P \wedge \sim Q$. Then, we analyzed each option by converting the implication into a disjunction using the implication equivalence and applying De Morgan's laws. After simplification, we found that only option (D) simplifies to a tautology (T). The original solution had errors in its analysis of the options.

Final Answer

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