Key Concepts and Formulas

• Implication: $P \Rightarrow Q \equiv \sim P \vee Q$
• De Morgan's Laws: $\sim(P \vee R) \equiv \sim P \wedge \sim R$ and $\sim(P \wedge R) \equiv \sim P \vee \sim R$
• Distributive Law: $(A \wedge B) \vee C \equiv (A \vee C) \wedge (B \vee C)$ and $(A \vee B) \wedge C \equiv (A \wedge C) \vee (B \wedge C)$

Step-by-Step Solution

Step 1: Convert the implications to their equivalent disjunction forms. We start with the given statement $(P \Rightarrow Q) \wedge (R \Rightarrow Q)$. We use the implication equivalence $P \Rightarrow Q \equiv \sim P \vee Q$ to rewrite the statement. $(P \Rightarrow Q) \wedge (R \Rightarrow Q) \equiv (\sim P \vee Q) \wedge (\sim R \vee Q)$

Step 2: Apply the distributive law. We apply the distributive law on the right-hand side: $(A \wedge B) \vee C \equiv (A \vee C) \wedge (B \vee C)$ or $(A \vee B) \wedge C \equiv (A \wedge C) \vee (B \wedge C)$. Here, let $A = \sim P$, $B = \sim R$ and $C = Q$. $(\sim P \vee Q) \wedge (\sim R \vee Q) \equiv (\sim P \wedge \sim R) \vee Q$

Step 3: Apply De Morgan's Law. We use De Morgan's Law, $\sim P \wedge \sim R \equiv \sim (P \vee R)$. $(\sim P \wedge \sim R) \vee Q \equiv \sim(P \vee R) \vee Q$

Step 4: Convert back to implication form. We use the implication equivalence $P \Rightarrow Q \equiv \sim P \vee Q$ in reverse. So, $\sim(P \vee R) \vee Q \equiv (P \vee R) \Rightarrow Q$. $\sim(P \vee R) \vee Q \equiv (P \vee R) \Rightarrow Q$

Common Mistakes & Tips

• Carefully apply De Morgan's Laws. Remember the difference between $\sim(P \vee Q)$ and $\sim P \vee \sim Q$.
• The implication rule $P \Rightarrow Q \equiv \sim P \vee Q$ is fundamental. Make sure you understand and can apply it in both directions.
• When using distributive law, pay close attention to the operators (AND, OR) to avoid errors.

Summary

We started with the given statement $(P \Rightarrow Q) \wedge (R \Rightarrow Q)$, converted the implications to their equivalent disjunction forms, applied the distributive law and De Morgan's Law, and then converted back to implication form. This process simplified the expression to $(P \vee R) \Rightarrow Q$. Therefore, $(P \Rightarrow Q) \wedge (R \Rightarrow Q)$ is logically equivalent to $(P \vee R) \Rightarrow Q$.

Final Answer The final answer is \boxed{(P \vee R) \Rightarrow Q}, which corresponds to option (D).