Question

The negation of \( (p \land (\sim q)) \lor (\sim p) \) is equivalent to:

• A. \( p \land (\sim q) \)
• B. \( p \land q \)
• C. \( p \land (q \land (\sim p)) \)
• D. \( p \lor (q \lor (\sim p)) \)

Hint:

To find the negation of logical expressions, apply De Morgan's laws carefully, then simplify using logical identities and set theory concepts if needed.

Answer 1

The Correct Option is B

To find the negation of the expression \( (p \land (\sim q)) \lor (\sim p) \) and show its equivalence to one of the given options, we can use De Morgan's laws and other logical equivalences.

Step 1: Expression Breakdown

The given expression is:

\((p \land (\sim q)) \lor (\sim p)\)

Step 2: Negation of the Expression

We need the negation of the entire expression:

\(\sim [(p \land (\sim q)) \lor (\sim p)]\)

Step 3: Apply De Morgan's Laws

According to De Morgan's Laws, the negation of a disjunction (OR, \(\lor\)) is the conjunction (AND, \(\land\)) of the negations:

\(\sim [A \lor B] = \sim A \land \sim B\)

For our expression, let:

\(A = (p \land (\sim q))\) and \(B = (\sim p)\)

Therefore, the negation becomes:

\(\sim [(p \land (\sim q)) \lor (\sim p)] = \sim (p \land (\sim q)) \land \sim (\sim p)\)

Step 4: Simplify Each Part

• \(\sim (\sim p)\) is equivalent to \(p\\)
• For \(\sim (p \land (\sim q))\), we use again De Morgan's Laws:
  • \(\sim (p \land (\sim q)) = \sim p \lor q\)

Step 5: Combine the Negated Parts

Combine the results using the AND operation:

\((\sim p \lor q) \land p\)

By rearranging, we get:

\((p \land q) \lor (p \land \sim p)\)

Since \(p \land \sim p\) is always false (a contradiction), it simplifies to:

\(p \land q\)

Thus, the negation of the given expression is equivalent to:

\(p \land q\)

Conclusion:

The correct answer is therefore \(p \land q\).