Mathematical Logic Class 12 Commerce Maths 1 Chapter 1 Exercise 1.9 Answers Maharashtra Board
Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.9 Questions and Answers.
Std 12 Maths 1 Exercise 1.9 Solutions Commerce Maths
Question 1.
Without using truth table, show that
(i) p ↔ q ≡ (p ∧ q) ∨ (~p ∧ ~q)
Solution:
LHS = p ↔ q
≡ (p → q) ∧ (q → p)
≡ (~p ∨ q) ∧ (~(q ∨ p) …..(Conditional Law)
≡ [~p ∧ (~(q ∨ p)] ∨ [q ∧ (~q ∨ p] …..(Distributive Law)
≡ [(~p ∧ ~q) ∨ (~p ∧ p)] ∨ [(q ∧ ~q) ∨ (q ∧ p)] ……(Distributive Law)
≡ [(~p ∧ ~q) ∨ c] ∨ [c ∨ (q ∧ p)] …..(Complement Law)
≡ (~p ∧ ~q) ∨ (q ∧ p) ……(Identity Law)
≡ (~p ∧ ~q) ∨ (p ∧ q) ……(Commutative Law)
≡ (p ∧ q) ∨ (~p∧ q) ……(Commutative Law)
≡ RHS.
(ii) p ∧ [~p ∨ q) ∨ (~q)] ≡ p
Solution:
LHS = p ∧ [(~p ∨ q) ∨ (~q)]
≡ p ∧ [~p ∨ (q ∨ ~q)] ……(Associative Law)
≡ p ∧ [~p ∨ t] …….(Complement Law)
≡ p ∧ t ……(Identity Law)
≡ p ……(Identity Law)
= RHS.
(iii) ~[(p ∧ q) → ~(q)] ≡ p ∧ q
Solution:
LHS = ~[(p ∧ q) → ~(~q)]
≡ (p ∧ q) ∧ ~(~q) ……(Negation of implication)
≡ (p ∧ q) ∧ q …..(Negation of negation)
≡ p ∧ (q ∧ q) …..(Associative Law)
≡ P ∧ q ……(Idempotent Law)
= RHS
(iv) ~r → ~(p ∧ q) ≡ [~(q → r)] → (~p)
Solution:
LHS = ~r → ~(p ∧ q)
≡ ~q → (~p ∨ ~q) ……(De Morgan’s Law)
≡ ~(~r) ∨ (~p ∨~q) …..(Conditional Law)
≡ r ∨ (~p ∨ ~q) …..(Involution Law)
≡ r ∨ ~q ∨ ~p …..(Commutative Law)
≡ (~q ∨ r) ∨ (~p) ……(Commutative Law)
≡ ~(q → r) ∨ (~p) ……(Conditional Law)
≡ ~(q → r) → (~p) ……(Conditional Law)
= RHS.
(v) (p ∨ q) → r ≡ (p → r) ∧ (q → r)
Solution:
LHS = (p ∨ q) → r
≡ ~(p → q) ∨ r ……..(Conditional Law)
≡ (~p ∧ ~q) ∨ r ……….(De Morgan’s Law)
≡ (~p ∨ r) ∧ (~q ∨ r) ………..(Distributive Law)
≡ (p → r) ∧ (q → r) …….(Conditional Law)
= RHS.
Question 2.
Using the algebra of statement, prove that:
(i) [p ∧ (q ∨ r)] ∨ [~r ∧ ~q ∧ p] ≡ p
Solution:
LHS = [p ∧ (q ∨ r)] ∨ [ ~r ∧ ~q ∧ p]
≡ [p ∧ (q ∨ r)] ∨ [(~r ∧ ~q) ∧ p] ……(Associative Law)
≡ [p ∧ (q ∨ r)] ∨ [(~q ∧ ~r) ∧ p] ……(Commutative Law)
≡ [p ∧ (q ∨ r)] ∨ [ ~ (q ∨ r) ∧ p] ……(De Morgan’s Law)
≡ [p ∧ (q ∨ r)] ∨ [p ∧ ~(q ∨ r)] ……(Commutative Law)
≡ p ∧ [(q ∨ r) ∨ ~ (q ∨ r) ] …..(Distributive Law)
≡ p ∧ t …….(Complement Law)
≡ p ……(Identity Law)
= RHS.
(ii) (p ∧ q) ∨ (p ∧ ~q) ∨ (~p ∧ ~q) ≡ p ∨ ~q
Solution:
LHS = (p ∧ q) ∨ (p ∧ ~q) ∨ (~p ∧ ~ q)
≡ (p ∧ q) ∨ [(p ∧ ~q) ∨ (~p ∧ ~q)] ……(Associative Law)
≡ (p ∧ q) ∨ [(~q ∧ p) ∨ (~q ∧ ~p)] …..(CommutativeLaw)
≡ (p ∧ q) ∨ [ ~q ∧ (p ∨ ~ p)] …..(Distributive Law)
≡ (p ∧ q) ∨ (~q ∧ t) ……(Complement Law)
≡ (p ∧ q) ∨ (~q) …….(Identity Law)
≡ (p ∨ ~q) ∧ (q ∨ ~q) ……(Distributive Law)
≡ (p ∨ ~q) ∧ t …….(Complement Law)
≡ p ∨ ~q …..(Identity Law)
= RHS.
(iii) (p ∨ q) ∧ (~p ∨ ~q) ≡ (p ∧ ~q) ∨ (~p ∧ q)
Solution:
LHS = (p ∨ q) ∧ (~p ∨ ~q)
≡ [p ∧ (~p ∨ ~q)] ∨ [q ∧ (~p ∨ ~q)] ……(Distributive Law)
≡ [(p ∧ ~p) ∨ (p ∧ ~q)] ∨ [q ∧ ~p) ∨ (q ∧ ~q)] ……(Distributive Law)
≡ [c ∨ (p ∧ ~q)] ∨ [(q ∧ ~p) ∨ c] ……(Complement Law)
≡ (p ∧ ~q) ∨ (q ∧ ~p) ……..(Identity Law)
≡ (p ∧ ~q) ∨ (~p ∧ q) ………(Commutative Law)
= RHS.