Question:medium

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

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Use Boolean algebra laws like De Morgan's theorem and distributive properties for simplification.
Updated On: Nov 26, 2025
  • \( (\sim p) \vee (\sim q) \)
  • \( p \vee (\sim q) \)
  • \( p \vee q \)
  • \( (\sim p) \vee q \)
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The Correct Option is A

Solution and Explanation

The objective is to simplify the provided Boolean expression step-by-step using Boolean algebra laws. The expression is: \[ (p \wedge \sim q) \vee ((\sim p) \wedge q) \vee ((\sim p) \wedge (\sim q)) \]
Step 1: Term Grouping Rearrange the expression for enhanced clarity by grouping terms: \[ (p \wedge \sim q) \vee \big[(\sim p \wedge q) \vee (\sim p \wedge \sim q)\big] \] 
Step 2: Distributive Law Application (Second Group) Within \((\sim p \wedge q) \vee (\sim p \wedge \sim q)\), factor out \(\sim p\): \[ (\sim p \wedge q) \vee (\sim p \wedge \sim q) = \sim p \wedge (q \vee \sim q) \] 
Applying the Complement Law, where \(q \vee \sim q = 1\): \[ \sim p \wedge (q \vee \sim q) = \sim p \wedge 1 \] 
Using the Identity Law, \(\sim p \wedge 1 = \sim p\): \[ (\sim p \wedge q) \vee (\sim p \wedge \sim q) = \sim p \]
Step 3: Substitution into Main Expression Substitute the simplified second group, \(\sim p\), back into the primary expression: \[ (p \wedge \sim q) \vee \sim p \
Step 4: Distributive Law Application Reorder and factor out \(\sim p\) from \((p \wedge \sim q) \vee \sim p\): \[ (p \wedge \sim q) \vee \sim p = \sim p \vee (p \wedge \sim q) \] This is equivalent to \(\sim p \vee (\sim q \wedge p)\) by the Distributive Law
Step 5: Absorption Law Application Applying the Absorption Law to \(\sim p \vee (\sim q \wedge p)\) yields \(\sim p \vee \sim q\). The expression is now simplified to: \[ \sim p \vee \sim q \] 

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

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