Question

The contrapositive of the statement pattern \([p∨(p\rightarrow q)]\rightarrow (p∧\sim q)\) is

• A. \((p∧\sim q)\rightarrow [p∧(p\rightarrow q)]\)
• B. \((\sim p∧\sim q)\rightarrow [\sim p∧(p\rightarrow \sim q)]\)
• C. \((\sim p∨q)∧[\sim p∨(p∧\sim q)]\)
• D. \((\sim p∨q)\rightarrow [\sim p∧(p∧\sim q)]\)

Hint:

The contrapositive of A implies B is not-B implies not-A.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
For a conditional \(A \to B\) the contrapositive is \(\sim B \to \sim A\). Here \(A = p \vee (p \to q)\) and \(B = p \wedge \sim q\).

Step 2: Key Formula or Approach:
De Morgan: \(\sim(p \wedge \sim q) \equiv \sim p \vee q\). And \(\sim(p \vee X) \equiv \sim p \wedge \sim X\).

Step 3: Detailed Explanation:
\(\sim B = \sim(p \wedge \sim q) = \sim p \vee q\).
\(\sim A = \sim[p \vee (p \to q)] = \sim p \wedge \sim(p \to q)\).
Since \(p \to q \equiv \sim p \vee q\), its negation is \(p \wedge \sim q\). So \(\sim A = \sim p \wedge (p \wedge \sim q)\).
\[ \sim B \to \sim A:\ (\sim p \vee q) \to [\sim p \wedge (p \wedge \sim q)] \]
This is option D. Options A and B do not have \(\sim p \vee q\) as the hypothesis, and C is not a conditional at all.

Final Answer:
The contrapositive is option (D). \[ \boxed{(\sim p \vee q) \to [\sim p \wedge (p \wedge \sim q)]} \]