Question

Let $p$ and $q$ be two statements Then $\sim(p \wedge(p \Rightarrow \sim q))$ is equivalent to

• A. $p \vee(p \wedge q)$
• B. $(\sim p) \vee q$
• C. $p \vee(p \wedge(\sim q))$
• D. $p \vee((\sim p) \wedge q)$

Answer 1

The Correct Option is B

The correct answer is (B) : $(\sim p) \vee q$
∼(p∧(p→∼q)) 
≡∼p∨∼(∼p∨∼q) 
≡∼p∨(p∧q) 
≡(∼p∨p)∧(∼p∨q) 
≡t∧(∼p∨q) 
≡∼p∨q

Answer 2

Step 1: Expand \( \sim (p \land (p \rightarrow \sim q)) \) 

First, expand \( p \rightarrow \sim q \):

\[ p \rightarrow \sim q \equiv \sim p \lor \sim q. \] So: \[ \sim (p \land (p \rightarrow \sim q)) \equiv \sim (p \land (\sim p \lor \sim q)). \] 

Step 2: Apply De Morgan's law

Using \( \sim (A \land B) \equiv \sim A \lor \sim B \):

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

Step 3: Simplify \( \sim (\sim p \lor \sim q) \)

Using \( \sim (A \lor B) \equiv \sim A \land \sim B \):

\[ \sim (\sim p \lor \sim q) \equiv p \land q. \] Substituting back into the expression: \[ \sim p \lor (p \land q). \] 

Step 4: Distribute \( \lor \)

Using the distributive laws of logic:

\[ \sim p \lor (p \land q) \equiv (\sim p \lor p) \land (\sim p \lor q). \] Since \( \sim p \lor p \equiv \text{True} \), we get: \[ \sim p \lor (p \land q) \equiv \sim p \lor q. \]