Key Concepts and Formulas

• De Morgan's Laws:
  • $\sim(p \vee q) \equiv (\sim p) \wedge (\sim q)$
  • $\sim(p \wedge q) \equiv (\sim p) \vee (\sim q)$
• Distributive Laws:
  • $p \wedge (q \vee r) \equiv (p \wedge q) \vee (p \wedge r)$
  • $p \vee (q \wedge r) \equiv (p \vee q) \wedge (p \vee r)$
• Negation of Negation: $\sim(\sim p) \equiv p$
• Commutative Laws:
  • $p \vee q \equiv q \vee p$
  • $p \wedge q \equiv q \wedge p$
• Associative Laws:
  • $(p \vee q) \vee r \equiv p \vee (q \vee r)$
  • $(p \wedge q) \wedge r \equiv p \wedge (q \wedge r)$
• Identity Laws:
  • $p \vee F \equiv p$
  • $p \wedge T \equiv p$
• Inverse Laws:
  • $p \vee (\sim p) \equiv T$
  • $p \wedge (\sim p) \equiv F$
• Absorption Laws:
  • $p \vee (p \wedge q) \equiv p$
  • $p \wedge (p \vee q) \equiv p$
• Conditional and Biconditional Statements: These are not strictly necessary for this question, but are good to know.

Step-by-Step Solution

Step 1: Apply De Morgan's Law to the outermost negation. We have the statement $\sim[p \vee (\sim(p \wedge q))]$. Applying De Morgan's Law, we distribute the negation and change the $\vee$ to $\wedge$: $\sim[p \vee (\sim(p \wedge q))] \equiv (\sim p) \wedge \sim(\sim(p \wedge q))$

Step 2: Simplify the double negation. We have $\sim(\sim(p \wedge q))$, which simplifies to $(p \wedge q)$. $(\sim p) \wedge \sim(\sim(p \wedge q)) \equiv (\sim p) \wedge (p \wedge q)$

Step 3: Apply the associative law of conjunction. We can regroup the terms using the associative law: $(\sim p) \wedge (p \wedge q) \equiv (\sim p \wedge p) \wedge q$

Step 4: Apply the inverse law. We know that $p \wedge (\sim p) \equiv F$. $(\sim p \wedge p) \wedge q \equiv F \wedge q$

Step 5: Apply the identity law. We know that $F \wedge q \equiv F$. $F \wedge q \equiv F$

Step 6: Analyze the options to find an equivalent expression. We need to find an option that is always false. Option (A): $(\sim(p \wedge q)) \wedge q$ $(\sim(p \wedge q)) \wedge q \equiv ((\sim p) \vee (\sim q)) \wedge q \equiv ((\sim p) \wedge q) \vee ((\sim q) \wedge q) \equiv ((\sim p) \wedge q) \vee F \equiv (\sim p) \wedge q$ This is not always false.

Option (B): $\sim(p \vee q) \equiv (\sim p) \wedge (\sim q)$. This is not always false.

Option (C): $(p \wedge q) \wedge (\sim p) \equiv (p \wedge (\sim p)) \wedge q \equiv F \wedge q \equiv F$. This is always false.

Option (D): $\sim(p \wedge q) \equiv (\sim p) \vee (\sim q)$. This is not always false.

Since our simplified expression is $F$, option (C) is equivalent.

However, the correct answer is stated to be option (A). Let's re-examine the initial problem and the target.

We simplified the original expression to $(\sim p) \wedge (p \wedge q)$. We want to show this is equivalent to $(\sim(p \wedge q)) \wedge q$.

$(\sim(p \wedge q)) \wedge q \equiv ((\sim p) \vee (\sim q)) \wedge q \equiv ((\sim p) \wedge q) \vee ((\sim q) \wedge q) \equiv ((\sim p) \wedge q) \vee F \equiv (\sim p) \wedge q$.

Now, let's see if $(\sim p) \wedge (p \wedge q) \equiv (\sim p) \wedge q$.

$(\sim p) \wedge (p \wedge q) \equiv ((\sim p) \wedge p) \wedge q \equiv F \wedge q \equiv F$. $(\sim p) \wedge q$ is not always F.

Let's analyze option (A) again: $(\sim(p \wedge q)) \wedge q$. $(\sim(p \wedge q)) \wedge q \equiv ((\sim p) \vee (\sim q)) \wedge q \equiv ((\sim p) \wedge q) \vee ((\sim q) \wedge q) \equiv ((\sim p) \wedge q) \vee F \equiv (\sim p) \wedge q$.

The original expression simplifies to $F$. We need to show $(\sim(p \wedge q)) \wedge q$ is also $F$. $(\sim(p \wedge q)) \wedge q \equiv (\sim p \vee \sim q) \wedge q \equiv (\sim p \wedge q) \vee (\sim q \wedge q) \equiv (\sim p \wedge q) \vee F \equiv \sim p \wedge q$. This is equal to T when p is F and q is T. So, option A is NOT correct.

There seems to be an error in the given answer. The original expression reduces to $F$, which is equivalent to option (C).

Common Mistakes & Tips

• Be careful when applying De Morgan's Laws. Remember to negate each term and change the connective.
• Always double-check your simplifications to avoid errors.
• When using truth tables, ensure you consider all possible combinations of truth values for the variables.

Summary

We started with the given expression $\sim[p \vee(\sim(p \wedge q))]$ and simplified it using De Morgan's Laws, the associative law, and inverse law to obtain $F$. Comparing this result with the options, we found that option (C), $(p \wedge q) \wedge(\sim p)$, also simplifies to $F$. Therefore, the given answer of option (A) is incorrect, and the correct answer should be option (C).

Final Answer

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