Question

The statement \( (p \wedge (\sim q)) \vee ((\sim p) \wedge q) \vee ((\sim p) \wedge (\sim q)) \) is equivalent to:

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

Hint:

Use Boolean algebra laws like De Morgan's theorem and distributive properties for simplification.

Answer 1

The Correct Option is A

We are tasked with simplifying the Boolean expression step by step using the laws of Boolean algebra. The given expression is: \[ (p \wedge \sim q) \vee ((\sim p) \wedge q) \vee ((\sim p) \wedge (\sim q)) \] 
Step 1: Group the terms We rewrite the expression for clarity by grouping terms: \[ (p \wedge \sim q) \vee \big[(\sim p \wedge q) \vee (\sim p \wedge \sim q)\big] \] 
Step 2: Apply the Distributive Law to the second group In \((\sim p \wedge q) \vee (\sim p \wedge \sim q)\), factor out \(\sim p\): \[ (\sim p \wedge q) \vee (\sim p \wedge \sim q) = \sim p \wedge (q \vee \sim q) \] 
Using the Complement Law, \(q \vee \sim q = 1\): \[ \sim p \wedge (q \vee \sim q) = \sim p \wedge 1 \] 
Using the Identity Law, \(\sim p \wedge 1 = \sim p\): \[ (\sim p \wedge q) \vee (\sim p \wedge \sim q) = \sim p \]
Step 3: Substitute back into the main expression Replace \((\sim p \wedge q) \vee (\sim p \wedge \sim q)\) with \(\sim p\): \[ (p \wedge \sim q) \vee \sim p \
Step 4: Apply the Distributive Law Factor out \(p\) from \((p \wedge \sim q) \vee \sim p\): \[ (p \wedge \sim q) \vee \sim p = \sim p \vee (p \wedge \sim q) \] By the Distributive Law, this is equivalent to: \[ \sim p \vee (\sim q \wedge p) \] 
Step 5: Apply the Absorption Law Using the Absorption Law, \(\sim p \vee (\sim q \wedge p) = \sim p \vee \sim q\). This simplifies the expression to: \[ \sim p \vee \sim q \] 

Final Simplified Expression: \[ \boxed{(\sim p) \vee (\sim q)} \]