Chapter 2 - The Logic of Compound Statements - Exercise Set 2.4 - Page 77: 30

Circuit will be

$(P ∧ Q) ∨ (∼P ∧ Q) ∨ (∼P∧ ∼Q)$ $≡ (P ∧ Q) ∨ ((∼P ∧ Q) ∨ (∼P∧ ∼Q))$ by inserting parentheses (which is legal by the associative law) $≡ (P ∧ Q) ∨ (∼P ∧ (Q∨ ∼Q))$ by the distributive law $≡ (P ∧ Q) ∨ (∼P ∧ t) $ by the negation law for ∨ ≡ $(P ∧ Q)∨ ∼P$ by the identity law for ∧ ≡$∼P ∨ (P ∧ Q)$ by the commutative law for ∨ ≡ $(∼P ∨ P) ∧ (∼P ∨ Q)$ by the distributive law ≡ $(P∨ ∼P) ∧ (∼P ∨ Q)$ by the commutative law for ∨ ≡ $t ∧ (∼P ∨ Q)$ by the negation law for ∨ ≡ $(∼P ∨ Q) ∧ t$ by the commutative law for ∧ ≡$∼P ∨ Q$ by the identity law for ∧