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Question No 1: Construct a truth table for each of these compound propositions. I. (p → q) ↔ (¬q → ¬p) p q ~p ~q (p → q) ~ (q→ ~p) (p → q) ↔ (¬q → ¬p) T T F F T T T T F F T F F T F T T F T T T F F T T T T T II. (p ↔ q) ⊕ (p ↔ ¬q) p q ~p ~q (p → q) ~ (q→ ~p) (p → q) ↔ (¬q → ¬p) T T F F T T T T F F T F F T F T T F T T T F F T T T T T P q ~q (p ↔ q) (p ↔ ¬q) (p ↔ q) ⊕ (p ↔ ¬q) T T F T F T
III. (p → q) ∧ (¬p → r) p q r ~P (p → q) (¬p → r) (p → q) ∧ (¬p → r) T T T F T T T T T F F T T T T F T F F T F T F F F F T F F T T T T T T F T F T T F F F F T T T T T F F F T T F F Question No. I. Determine whether (¬q ∧ (p → q)) → ¬p is a tautology. p q ~p ~q (p → q) ¬q ∧ (p → q) (¬q ∧ (p → q)) → ¬p T T F F T F T T F F T F F T F T T F T F T
II. Using Laws of logic Prove: ¬p↔ q ≡ p ↔¬q III. Without using Truth Table Show that q∧¬ (p→q) is a contradiction
p → q≡~(p∧~q) Implication Law q∧~(~(p∧~q)) q∧(p∧~q) Double Negation Law q ∧ (~q ∧ p) Commutative Law (q∧~q) ∧p Associative Law = p∧c Universal Bound Law = c proved IV. ¬ (p ˅ (¬p˄q) ≡¬p˄ (¬p˄q) (¬p q) ≡¬p˄q) ≡¬p˄ (¬p˄q) ˄q) ≡¬p˄ (¬p˄q) (¬p q)˄q) ≡¬p˄ (¬p˄q) V. (s→ r) ˄q) ≡¬p˄ (¬p˄q) (q → r) ≡ (s ˅ (¬p˄q) ≡¬p˄ (¬p˄q)q) → r
