Question

Without using truth table prove that \( (p \wedge q) \vee (\sim p \wedge q) \vee (p \wedge \sim q) \equiv p \vee q \)

Hint:

Use distributive and absorption laws for logical equivalences; group terms to simplify.

Answer 1

Step 1: Group the first two terms:
\[ (p \wedge q) \vee (\sim p \wedge q) = q \wedge (p \vee \sim p) = q \wedge T = q. \]

Step 2: Now the expression becomes:
\[ q \vee (p \wedge \sim q). \]

Step 3: Distribute using absorption law:
\[ q \vee (p \wedge \sim q) = (q \vee p) \wedge (q \vee \sim q) = (p \vee q) \wedge T = p \vee q. \] Thus, \( (p \wedge q) \vee (\sim p \wedge q) \vee (p \wedge \sim q) \equiv p \vee q \).