Mathematical Logic Exercise 1.9 [Page 22] Balbharati solutions for Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board Chapter 1

EXERCISE 1.9 [PAGE 22]

EXERCISE 1.9PAGE 22

Balbharati solutions for Mathematics and Statistics 1 (Commerce) 12th Standard HSC Maharashtra State Board Chapter 1 Mathematical Logic Exercise 1.9 [Page 22]

EXERCISE 1.9Q 1.1   PAGE 22

Exercise 1.9 | Q 1.1 | Page 22

Without using truth table, show that

p ↔ q ≡ (p ∧ q) ∨ (~p ∧ ~q)

SOLUTION

L.H.S.

≡ p ↔ q

≡ (p → q) ∧ (q → p)

≡ (~p ∨ q) ∧ (~q ∨ p)

≡ [~ p ∧ (~ q ∨ p)] ∨ [q ∧ (~ q ∨ p)]     ....[Distributive law]

≡ [(~ p ∧ ~ q) ∨ (~ p ∧ p)] ∨ [(q ∧ ~ q) ∨ (q ∧ p)]     .....[Distributive Law]

≡ [(~ p ∧ ~ q) ∨ F] ∨ [F ∨ (q ∧ p)]  ....[Complement Law]

≡ (~ p ∧ ~ q) ∨ (q ∧ p)   ....[Identity Law]

≡ (p ∧ q) ∨ (~ p ∧ ~ q)    ....[Commutative Law]

≡ R.H.S.

EXERCISE 1.9Q 1.2   PAGE 22

Exercise 1.9 | Q 1.2 | Page 22

Without using truth table, show that

p ∧ [(~ p ∨ q) ∨ ~ q] ≡ p

SOLUTION

L.H.S.

≡ p ∧ [(~ p ∨ q) ∨ ~ q]

≡ p ∧ [(~ p ∨ (q ∨ ~ q)]     .....[Associative law]

≡ p ∧ (~ p ∨ T)       .....[Complement law]

≡ p ∧ T                 .....[Identity law]

≡ p                       .....[Identity law] 

≡ R.H.S.

EXERCISE 1.9Q 1.3   PAGE 22

Exercise 1.9 | Q 1.3 | Page 22

Without using truth table, show that

~ [(p ∧ q) → ~ q] ≡ p ∧ q

SOLUTION

L.H.S.

≡ ~ [(p ∧ q) → ~ q]

≡ (p ∧ q) ∧ ~ (~ q)   ....[Negation of implication]

≡ (p ∧ q) ∧ q      .....[Negation of a negation]

≡ p ∧ (q ∧ q)     ....[Associative law]

≡ p ∧ q          .....[Identity law]

≡ R.H.S.

EXERCISE 1.9Q 1.4   PAGE 22

Exercise 1.9 | Q 1.4 | Page 22

Without using truth table, show that

~r → ~ (p ∧ q) ≡ [~ (q → r)] → ~ p

SOLUTION

L.H.S.

≡ ~r → ~ (p ∧ q)

≡ ~(~ r) ∨ ~ (p ∧ q)       ....[p → q ≡ ~ p ∨ q]

≡ r ∨ ~(p ∧ q)             ....[Negation of negation]

≡ r ∨ (~p ∨ ~q)           ....[De Morgan’s law]

≡ ~p ∨ (~q ∨ r)         .....[Commutative and associative law]

≡ ~p ∨ (q → r)          ....[p → q ≡ ~ p ∨ q]

≡ (q → r) ∨ ~p            ......[Commutative law]

≡ ~[~ (q → r)] ∨ ~ p     ......[Negation of negation]

≡ [~ (q → r)] → ~ p        .....[p → q ≡ ~ p ∨ q]

= R.H.S.

EXERCISE 1.9Q 1.5   PAGE 22

Exercise 1.9 | Q 1.5 | Page 22

Without using truth table, show that

(p ∨ q) → r ≡ (p → r) ∧ (q → r)

SOLUTION

L.H.S.

≡ (p ∨ q) → r

≡ ~ (p ∨ q) ∨ r          ....[p → q → ~ p ∨ q]

≡ (~ p ∧ ~ q) ∨ r       ....[De Morgan’s law] 

≡ (~ p ∨ r) ∧ (~ q ∨ r)       .....[Distributive law]

≡ (p → r) ∧ (q → r)             .....[p → q → ~ p ∨ q]

= R.H.S.

EXERCISE 1.9Q 2.1   PAGE 22

Exercise 1.9 | Q 2.1 | Page 22

Using the algebra of statement, prove that

[p ∧ (q ∨ r)] ∨ [~ r ∧ ~ q ∧ p] ≡ p

SOLUTION

L.H.S.

= [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]

= R.H.S.

EXERCISE 1.9Q 2.2   PAGE 22

Exercise 1.9 | Q 2.2 | Page 22

Using the algebra of statement, prove that

(p ∧ q) ∨ (p ∧ ~ q) ∨ (~ p ∧ ~ q) ≡ (p ∨ ~ q)

SOLUTION

L.H.S.

= (p ∧ q) ∨ (p ∧ ~ q) ∨ (~ p ∧ ~ q)

≡ (p ∧ q) ∨ [(p ∧ ~ q) ∨ (~ p ∧ ~ q)]    ....[Associative Law]

≡ (p ∧ q) ∨ [(~q ∧ p) ∨ (~ q ∧ ~ p)]    ....[Commutative Law]

≡ (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]

= R.H.S.

EXERCISE 1.9Q 2.3   PAGE 22

Exercise 1.9 | Q 2.3 | Page 22

Using the algebra of statement, prove that

SOLUTION

(p ∨ q) ∧ (~ p ∨ ~ q) ≡ (p ∧ ~ q) ∨ (~ p ∧ q)

L.H.S.

= (p ∨ q) ∧ (~ p ∨ ~ q)

≡ [(p ∨ q) ∧ ~ p] ∨ [(p ∨ q) ∧ ~ q]     .....[Distributive law]

≡ [(p ∧ ~ p) ∨ (q ∧ ~ p)] ∨ [(p ∧ ~ q) ∨ (q ∧ ~ q)]        .....[Distributive law]

≡ [F ∨ (q ∧ ~p)] ∨ [(p ∧ ~ q) ∨ F]        .....[Complement law]

≡ (q ∧ ~p) ∨ (p ∧ ~ q)      .....[Identity law]

≡ (p ∧ ~ q) ∨ (~ p ∧ q)      ....[Commutative law]

= R.H.S.