1. Assignment 2Deadline: April 5 (Thursday Class), April 6(Friday Class) Harshit Kumar April 25, 2012
2. Q1. Which of the following are proposition? (You answer should be T/F.) 1. y < 20. - F 2. It is sunny. - T 3. open the closet. - F, (Because, it is an order) 4. 10 > 5. T, (In fact, it is a tautology. Tautology is something which is always true.) 5. a2 + b2 = c2 . - F Q2. Let H=”Kim is smart”, C=”Kim is cunning”, O=”Kim is intelligent”. Rewrite the following using the logical operators ∧, ∨, and ¬ 1. Kim is smart and cunning but not intelligent. H ∧ C ∧ ¬O 2. Kim is either smart or cunning or both. H ∨ C 3. Kim is either smart or cunning but not both. (H ∨ C) ∧ ¬(H ∧ C) 4. Kim is neither smart, cunning nor intelligent. ¬(H ∨ C ∨ O) 5. Kim is not both smart and cunning, but he is intelligent.¬(H ∧ C) ∧ O 6. Kim is intelligent but not cunning nor smart. O ∧ ¬(C ∨ H) Q3. Simplify the following logical formula 1. p ⊕ q - (p ∧ ¬q) ∨ (¬p ∨ q) 2. p → ¬q - ¬p ∧ ¬q 3. (¬p ↔ ¬q) ∧ q - ¬p ∧ q 4. ¬(p → q) - p ∧ ¬q Q4. Construct logical formula for the truth table below. After constructing the logical formula, simplify it.a. a b c f(a,b,c) T T T T T T F F T F T F T F F F F T T F F T F F F F T T F F F T Sol. f (a, b, c) = (a ∧ b ∧ c) ∨ (¬a ∧ ¬b).b. a b c f(a,b,c) T T T T T T F F T F T T T F F T F T T T F T F F F F T T F F F F Sol. (¬b ∧ a) ∨ c.Q5. Prove the equivalence of following logical formula? You answer should be T/F. You answer should also show why it is T/F. Show each step. 1. (P ∧ Q) ∨ R and P ∧ (Q ∨ R). - F 2. ¬(P ⊕ Q) and P ↔ Q. - T 3. ¬(P ∨ Q ∨ R) and ¬P ∧ ¬Q ∧ ¬R. - T 4. P ∧ (P ∨ Q) and P ∨ (P ∧ Q). - T 5. (P ∧ Q) ∨ (Q ∧ R) and Q ∨ (P ∧ R). - F Q6. Simplify the following logical formula. 1. ¬((¬P ∧ (¬Q ∨ P )) ∨ ¬R). - (P ∨ Q) ∧ R 2. (¬P ∧ ¬(P ∧ R)) ∨ ((Q ∨ (Q ∧ R)) ∧ (Q ∨ S)). - ¬P ∨ q 3. ((¬P ∧ Q) ∧ (Q ∧ R)) ∧ ¬Q. - F 4. ¬(¬Q ∧ ¬(¬Q ∨ S)) ∨ (Q ∧ (r → r)). - T 5. ¬P ∧ (P ∨ Q) ∨ (Q ∨ (P ∧ P )) ∧ (P ∨ ¬Q). - P ∨ QQ7. Simplify the following logical formula. 1. P → Q. - ¬p ∨ q 2. ¬Q → ¬P. - ¬p ∨ q 3. Q → P. - p ∨ ¬q 4. ¬P → ¬Q. - p ∨ ¬q 1
3. 5. ¬(P → Q.) - p ∧ ¬q Q8. Prove the following using resolution. 1. ((A → B) → C) → (¬C → A) 2. ¬(¬P ∨ ¬Q) → ((P → R) ∧ (Q → R))Ans 8.1 2
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