Key Concepts and Formulas

• Distributive Law: $p \wedge (q \vee r) \equiv (p \wedge q) \vee (p \wedge r)$ and $p \vee (q \wedge r) \equiv (p \vee q) \wedge (p \vee r)$
• Complement Law: $p \vee \sim p \equiv t$ (tautology) and $p \wedge \sim p \equiv f$ (contradiction)
• Identity Law: $p \wedge t \equiv p$ and $p \vee f \equiv p$
• Commutative Law: $p \wedge q \equiv q \wedge p$ and $p \vee q \equiv q \vee p$
• Associative Law: $(p \wedge q) \wedge r \equiv p \wedge (q \wedge r)$ and $(p \vee q) \vee r \equiv p \vee (q \vee r)$
• Absorption Law: $p \vee (p \wedge q) \equiv p$ and $p \wedge (p \vee q) \equiv p$
• De Morgan's Laws: $\sim(p \wedge q) \equiv \sim p \vee \sim q$ and $\sim(p \vee q) \equiv \sim p \wedge \sim q$

Step-by-Step Solution

Step 1: Rewrite the given expression. We are given the expression $((p \wedge q) \vee (p \vee \sim q)) \wedge (\sim p \wedge \sim q)$. Our goal is to simplify this expression using Boolean algebra laws.

Step 2: Apply the distributive law to the first part of the expression. The original solution incorrectly attempts to apply the distributive law in a non-standard way. Instead, we'll use the absorption law later. Keep the original expression as it is for now: $((p \wedge q) \vee (p \vee \sim q)) \wedge (\sim p \wedge \sim q)$

Step 3: Apply the associative law to the first part of the expression. $((p \wedge q) \vee (p \vee \sim q)) \wedge (\sim p \wedge \sim q) \equiv (p \vee (p \wedge q) \vee \sim q) \wedge (\sim p \wedge \sim q)$ Here, we rearranged the terms using the associative and commutative laws.

Step 4: Apply the absorption law. $p \vee (p \wedge q) \equiv p$. Thus, $(p \vee (p \wedge q) \vee \sim q) \wedge (\sim p \wedge \sim q) \equiv (p \vee \sim q) \wedge (\sim p \wedge \sim q)$.

Step 5: Apply the distributive law. $(p \vee \sim q) \wedge (\sim p \wedge \sim q) \equiv (p \wedge (\sim p \wedge \sim q)) \vee (\sim q \wedge (\sim p \wedge \sim q))$.

Step 6: Apply the associative law. $(p \wedge (\sim p \wedge \sim q)) \vee (\sim q \wedge (\sim p \wedge \sim q)) \equiv ((p \wedge \sim p) \wedge \sim q) \vee ((\sim q \wedge \sim q) \wedge \sim p)$.

Step 7: Apply the complement law and idempotent law. Since $p \wedge \sim p \equiv f$ and $\sim q \wedge \sim q \equiv \sim q$, we have: $(f \wedge \sim q) \vee (\sim q \wedge \sim p)$.

Step 8: Apply the identity law and commutative law. Since $f \wedge \sim q \equiv f$, we have: $f \vee (\sim q \wedge \sim p) \equiv \sim q \wedge \sim p$. By the commutative law, this is equivalent to $\sim p \wedge \sim q$.

Step 9: Use De Morgan's law. $\sim p \wedge \sim q \equiv \sim (p \vee q)$. This is still not the correct answer, which means there's an error in the original problem statement or the provided correct answer. Let's review the steps.

Step 10: Re-examine the target answer The correct answer is supposed to be $p \wedge q$. Let's try to prove that the original expression is equivalent to $p \wedge q$ is FALSE. If we consider $p = T$ and $q = F$, the original expression becomes $((T \wedge F) \vee (T \vee \sim F)) \wedge (\sim T \wedge \sim F) \equiv (F \vee (T \vee T)) \wedge (F \wedge T) \equiv (F \vee T) \wedge F \equiv T \wedge F \equiv F$. However, $p \wedge q = T \wedge F = F$. So this case is consistent.

Let's try $p = F$ and $q = F$. The original expression is $((F \wedge F) \vee (F \vee \sim F)) \wedge (\sim F \wedge \sim F) \equiv (F \vee (F \vee T)) \wedge (T \wedge T) \equiv (F \vee T) \wedge T \equiv T \wedge T \equiv T$. However, $p \wedge q = F \wedge F = F$. This is a CONTRADICTION. Therefore, the correct answer cannot be $p \wedge q$.

Let's assume the correct answer is actually $\sim p \wedge \sim q$ as we derived.

Common Mistakes & Tips

• Be careful when applying the distributive law. Make sure you distribute correctly.
• Remember the absorption law: $p \vee (p \wedge q) \equiv p$ and $p \wedge (p \vee q) \equiv p$. This can simplify expressions significantly.
• When simplifying Boolean expressions, it is often helpful to use truth tables to verify your results.

Summary

We started with the given Boolean expression and applied various Boolean algebra laws, including the associative, commutative, distributive, complement, identity, and absorption laws. We found that the original expression simplifies to $\sim p \wedge \sim q$. However, we also proved that the correct answer $p \wedge q$ provided in the problem statement is incorrect.

Final Answer

The expression simplifies to $\sim p \wedge \sim q$, which corresponds to option (D). The final answer is \boxed{(\sim p) \wedge (\sim q)}.