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

We start with the given expression:

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

Apply the negation:

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

Using De Morgan's laws:

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

Here, \( A = (p \land (\sim q)) \) and \( B = (\sim p) \):

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

Simplify each term:

1. For \( \sim (p \land (\sim q)) \):
   Using De Morgan's laws:
   \[ \sim (p \land (\sim q)) = (\sim p) \lor q. \]
2. For \( \sim (\sim p) \):
   \[ \sim (\sim p) = p. \]

Thus, the expression becomes:

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

Distribute \( p \):

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

Since \( (\sim p) \land p = \text{False} \):

\[ \text{False} \lor (q \land p) = (q \land p). \]

Hence, the negation simplifies to:

\[ p \land q. \]