Question

The compound statement $(P \lor Q) \land (\sim P) \Rightarrow Q$ is equivalent to :

• A. $\sim(P \Rightarrow Q)$
• B. $P \land \sim Q$
• C. $\sim(P \Rightarrow Q) \Leftrightarrow P \land \sim Q$
• D. $P \lor Q$

Hint:

A statement of the form $A \Leftrightarrow A$ is always a tautology. When a given statement simplifies to {True}, look for an option that is also always true.

Answer 1

The Correct Option is C

Given compound statement: \[ (P \lor Q) \land (\sim P) \Rightarrow Q \] --- Step 1: Simplify the antecedent
\[ (P \lor Q) \land (\sim P) \] Using distributive law: \[ (P \land \sim P) \lor (Q \land \sim P) \] Since $P \land \sim P$ is always false: \[ (P \lor Q) \land (\sim P) \equiv Q \land \sim P \] --- Step 2: Rewrite the implication
\[ (Q \land \sim P) \Rightarrow Q \] Using the identity: \[ A \Rightarrow B \equiv \sim A \lor B \] \[ (Q \land \sim P) \Rightarrow Q \equiv \sim(Q \land \sim P) \lor Q \] --- Step 3: Apply De Morgan’s Law
\[ \sim(Q \land \sim P) = \sim Q \lor P \] So, \[ (\sim Q \lor P) \lor Q \] Rearranging: \[ P \lor (\sim Q \lor Q) \] Since $\sim Q \lor Q$ is a tautology (T): \[ P \lor T \equiv T \] --- Conclusion so far: The given statement is a tautology. --- Step 4: Check the options
Option (A): \[ \sim(P \Rightarrow Q) \equiv \sim(\sim P \lor Q) \equiv P \land \sim Q \] Not a tautology ❌ Option (B): \[ P \land \sim Q \] Not a tautology ❌ Option (D): \[ P \lor Q \] Not a tautology ❌ Option (C): \[ \sim(P \Rightarrow Q) \Leftrightarrow (P \land \sim Q) \] But we already know: \[ \sim(P \Rightarrow Q) \equiv P \land \sim Q \] So this becomes: \[ A \Leftrightarrow A \] which is always true (a tautology) ✔️ \[ \boxed{\text{Correct Answer: (C)}} \]