Key Concepts and Formulas

• De Morgan's Laws:
  • $\sim (A \wedge B) \equiv \sim A \vee \sim B$
  • $\sim (A \vee B) \equiv \sim A \wedge \sim B$
• Law of Double Negation: $\sim (\sim A) \equiv A$
• Commutative Laws:
  • $A \wedge B \equiv B \wedge A$
  • $A \vee B \equiv B \vee A$
• Associative Laws:
  • $(A \wedge B) \wedge C \equiv A \wedge (B \wedge C)$
  • $(A \vee B) \vee C \equiv A \vee (B \vee C)$
• Identity Laws:
  • $A \wedge T \equiv A$
  • $A \vee F \equiv A$
• Inverse Laws:
  • $A \wedge \sim A \equiv F$
  • $A \vee \sim A \equiv T$
• Distributive Laws:
  • $A \wedge (B \vee C) \equiv (A \wedge B) \vee (A \wedge C)$
  • $A \vee (B \wedge C) \equiv (A \vee B) \wedge (A \vee C)$
• Absorption Laws:
  • $A \vee (A \wedge B) \equiv A$
  • $A \wedge (A \vee B) \equiv A$

Step-by-Step Solution

Step 1: Write the given statement and its negation.

We are given the statement $p \wedge (q \wedge \sim(p \wedge q))$. We want to find the negation of this statement, which is $\sim [p \wedge (q \wedge \sim(p \wedge q))]$.

Step 2: Apply De Morgan's Law to the outer negation.

Using De Morgan's Law, $\sim (A \wedge B) \equiv \sim A \vee \sim B$, we get: $\sim [p \wedge (q \wedge \sim(p \wedge q))] \equiv \sim p \vee \sim (q \wedge \sim(p \wedge q))$

Step 3: Apply De Morgan's Law to the inner negation.

Applying De Morgan's Law again, $\sim (A \wedge B) \equiv \sim A \vee \sim B$, to the term $\sim (q \wedge \sim(p \wedge q))$, we get: $\sim (q \wedge \sim(p \wedge q)) \equiv \sim q \vee \sim (\sim(p \wedge q))$

Step 4: Simplify using the Law of Double Negation.

Using the Law of Double Negation, $\sim (\sim A) \equiv A$, we can simplify $\sim (\sim(p \wedge q))$ to $(p \wedge q)$. Therefore, $\sim q \vee \sim (\sim(p \wedge q)) \equiv \sim q \vee (p \wedge q)$

Step 5: Substitute back into the main expression.

Substituting this back into the expression from Step 2, we have: $\sim p \vee (\sim q \vee (p \wedge q))$

Step 6: Rearrange using the commutative law.

Using the commutative law for disjunction, we can rearrange the terms within the parenthesis: $\sim p \vee ( (p \wedge q) \vee \sim q) \equiv \sim p \vee ( \sim q \vee (p \wedge q))$

Step 7: Use the distributive law.

We have $\sim p \vee (\sim q \vee (p \wedge q))$. We want to show that this is equivalent to $\sim(p \wedge q) \vee p$. Note that $\sim(p \wedge q) = \sim p \vee \sim q$. Thus, we want to show that $\sim p \vee (\sim q \vee (p \wedge q)) = (\sim p \vee \sim q) \vee p$. Using the associative law for disjunction, $\sim p \vee (\sim q \vee (p \wedge q)) = (\sim p \vee \sim q) \vee (p \wedge q)$. Thus, we want to show that $(\sim p \vee \sim q) \vee (p \wedge q) = (\sim p \vee \sim q) \vee p$. This is not immediately obvious.

Let's consider the target expression first: $(\sim(p \wedge q)) \vee p \equiv (\sim p \vee \sim q) \vee p$. Using the associative law, we can rewrite this as $\sim p \vee (\sim q \vee p)$. Using the commutative law, we can rewrite this as $\sim p \vee (p \vee \sim q)$. Using the associative law again, we can rewrite this as $(\sim p \vee p) \vee \sim q$. Since $\sim p \vee p \equiv T$, we have $T \vee \sim q \equiv T$.

Now let's simplify the original expression $\sim p \vee (\sim q \vee (p \wedge q))$. Using the distributive law, we have: $\sim q \vee (p \wedge q) = (\sim q \vee p) \wedge (\sim q \vee q) = (\sim q \vee p) \wedge T = \sim q \vee p = p \vee \sim q$. Therefore, $\sim p \vee (\sim q \vee (p \wedge q)) = \sim p \vee (p \vee \sim q) = (\sim p \vee p) \vee \sim q = T \vee \sim q = T$. This means the negation is a tautology. However, this doesn't match any of the options.

Let's try to simplify $\sim p \vee (\sim q \vee (p \wedge q))$ differently. $\sim p \vee (\sim q \vee (p \wedge q)) \equiv \sim p \vee ((\sim q \vee p) \wedge (\sim q \vee q)) \equiv \sim p \vee ((\sim q \vee p) \wedge T) \equiv \sim p \vee (\sim q \vee p) \equiv \sim p \vee (p \vee \sim q) \equiv (\sim p \vee p) \vee \sim q \equiv T \vee \sim q \equiv T$.

We need to arrive at $(\sim(p \wedge q)) \wedge q$ $\sim(p \wedge (q \wedge \sim(p \wedge q))) = \sim p \vee \sim (q \wedge \sim(p \wedge q)) = \sim p \vee (\sim q \vee (p \wedge q)) = (\sim p \vee \sim q) \vee (p \wedge q) = \sim (p \wedge q) \vee (p \wedge q)$. But $A \vee \sim A = T$.

Let's re-examine the expression $p \wedge (q \wedge \sim(p \wedge q))$. $p \wedge (q \wedge \sim(p \wedge q)) = p \wedge (q \wedge (\sim p \vee \sim q)) = p \wedge ((q \wedge \sim p) \vee (q \wedge \sim q)) = p \wedge ((q \wedge \sim p) \vee F) = p \wedge (q \wedge \sim p) = (p \wedge q) \wedge \sim p = (p \wedge \sim p) \wedge q = F \wedge q = F$. Then the negation is $\sim F = T$.

Let us look at the options. (A) $(\sim(p \wedge q)) \wedge q = (\sim p \vee \sim q) \wedge q = (\sim p \wedge q) \vee (\sim q \wedge q) = (q \wedge \sim p) \vee F = q \wedge \sim p$. (B) $(\sim(p \wedge q)) \vee p = (\sim p \vee \sim q) \vee p = (\sim p \vee p) \vee \sim q = T \vee \sim q = T$. (C) $p \vee q$ (D) $\sim(p \vee q) = \sim p \wedge \sim q$

Since we determined that the negation is equivalent to $T$, option (B) is the correct answer.

The correct answer is $(\sim(p \wedge q)) \vee p$.

Common Mistakes & Tips

• Carefully apply De Morgan's Laws, paying attention to the order of operations.
• Remember the Law of Double Negation and how it simplifies expressions.
• When simplifying, consider using the distributive, associative, and commutative laws to rearrange terms and potentially reveal simpler forms.

Summary

The problem requires us to find the negation of a given logical expression. We applied De Morgan's Laws, the Law of Double Negation, and other logical equivalences to simplify the expression. By carefully manipulating the expression and comparing it to the given options, we arrived at the correct answer. The initial simplification led to a tautology, which then was compared to the options to find the equivalent statement.

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