# Mathematical Logic - Ex 1.4 | Maharashtra Board 12th Maths Solutions Chapter 1

### Mathematical Logic - Ex 1.4 | Maharashtra Board 12th Maths Solutions Chapter 1

Question 1.
Using rules of negation write the negations of the following with justification.

(i) ~q → p
Solution:
The negation of ~q → p is
~(~q → p) ≡ ~ q ∧ ~p…. (Negation of implication)

(ii) p ∧ ~q
Solution:
The negation of p ∧ ~q is
~(p ∧ ~q) ≡ ~p ∨ ~(~q) … (Negation of conjunction)
≡ ~ p ∨ q … (Negation of negation)

(iii) p ∨ ~q
Solution:
The negation of p ∨ ~ p is
~ (p ∨ ~(q) ≡ ~p ∧ ~(~(q) … (Negation of disjunction)
≡ ~ p ∧ q … (Negation of negation)

(iv) (p ∨ ~q) ∧ r
Solution:
The negation of (p ∨ ~ q) ∧ r is
~[(p ∨ ~q) ∧ r] ≡ ~(p ∨ ~q) ∨ ~r … (Negation of conjunction)
≡ [ ~p ∧ ~(~q)] ∨ ~ r… (Negation of disjunction)
≡ (~ p ∧ q) ∧ ~ r … (Negation of negation)

(v) p → (p ∨ ~q)
Solution:
The negation of p → (p ∨ ~q) is
~ [p → (p ∨ ~q)] ≡ p ∧ ~ (p ∧ ~p) … (Negation of implication)
≡ p ∧ [ ~ p ∧ ~ (~(q)] … (Negation of disjunction)
≡ p ∧ (~ p ∧ q) (Negation of negation)

(vi) ~(p ∧ q) ∨ (p ∨ ~q)
Solution:
The negation of ~(p ∧ q) ∨ (p ∨ ~q) is
~[~(p ∧ q) ∨ (p ∨ ~q)] ≡ ~[~(p ∧ q)] ∧ ~(p ∨ ~q) … (Negation of disjunction)
≡ ~[~(p ∧ q)] ∧ [ p ∧ ~(~q)] … (Negation of disjunction)
≡ (p ∧ q) ∧ (~ p ∧ q) … (Negation of negation)

(vii) (p ∨ ~q) → (p ∧ ~q)
Solution:
The negation of (p ∨ ~q) → (p ∧ ~q) is
~[(p ∨ ~q) → (p ∧ ~q)]
≡ (p ∨ ~q) ∧ ~(p ∧ ~q) … (Negation of implication)
≡ (p ∨ ~q) ∧ [ ~p ∨ ~(~q)] … (Negation of conjunction)
≡ (p ∨ ~q) ∧ (~p ∨ q) … (Negation of negation)

(viii) (~ p ∨ ~q) ∨ (p ∧ ~q)
Solution:
The negation of (~ p ∨ ~q) ∨ (p ∧ ~ q) is
~ [(~p ∨ ~q) ∨ (p ∧ ~ q)]
≡ ~(~p ∨ ~q) ∧ ~(p ∧ ~q) … (Negation of disjunction)
≡ [~(~p) ∧ ~(~q)] ∧ [~p ∨ ~(~q)] … (Negation of disjunction and conjunction)
≡ (p ∧ q) ∧ (~p ∨ q) … (Negation of negation)

### Question 2.Rewrite the following statements without using if .. then.

(i) If a man is a judge then he is honest.
Solution:
Since p → ≡ ~p ∨ q, the given statements can be written as :
A man is not a judge or he is honest.

(ii) It 2 is a rational number then 2–√ is irrational number.
Solution:
2 is not a rational number or 2–√ is irrational number.

(iii) It f(2) = 0 then f(x) is divisible by (x – 2).
Solution:
f(2) ≠ 0 or f(x) is divisible by (x – 2).

### Question 3.Without using truth table prove that :

(i) p ↔ q ≡ (p∧ q) ∨ (~ p ∧ ~q)
Solution:
LHS = p ↔ q
≡ (p ↔ q) ∧ (q ↔ p) … (Biconditional Law)
≡ (~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) ∨ F] ∨ [F ∨ (q ∧ p)] … (ComplementLaw)
≡ (~ p ∧ ~ q) ∨ (q ∧ p) … (Identity Law)
≡ (~ p ∧ ~ q) ∨ (p ∧ q) … (Commutative Law)
≡ (p ∧ q) ∨ (~p ∧ ~q) … (Commutative Law)
≡ RHS.

(ii) (p ∨ q) ∧ (p ∨ ~q) ≡ p
Solution:
LHS = (p ∨ q) ∧ (p ∨ ~q)
≡ p ∨ (q ∧ ~q) … (Distributive Law)
≡ p ∨ F … (Complement Law)
≡ p … (Identity Law)
≡ RHS.

(iii) (p ∧ q) ∨ (~ p ∧ q) ∨ (p ∧ ~q) ≡ p ∨ q
Solution:
LHS = (p ∧ q) v (~p ∧ q) ∨ (p ∧ ~q)
≡ [(p ∨ ~p) ∧ q] ∨ (p ∧ ~q) … (Distributive Law)
≡ (T ∧ q) ∨ (p ∧ ~q) … (Complement Law)
≡ q ∨ (p ∧ ~q) … (Identity Law)
≡ (q ∨ p) ∧ (q ∨ ~q) … (Distributive Law)
≡ (q ∨ p) ∧ T .. (Complement Law)
≡ q ∨ p … (Identity Law)
≡ p ∨ q … (Commutative Law)
≡ RHS.

(iv) ~[(p ∨ ~q) → (p ∧ ~q)] ≡ (p ∨ ~q) ∧ (~p ∨ q)
Solution:
LHS = ~[(p ∨ ~q) → (p ∧ ~q)]
≡ (p ∨ ~q) ∧ ~(p ∧ ~q) … (Negation of implication)
≡ (p ∨ ~q) ∧ [~p ∨ ~(~q)] … (Negation of conjunction)
≡ (p ∨ ~ q) ∧ (~p ∨ q)… (Negation of negation)
≡ RHS.

### Mathematical Logic - Ex 1.4 | Maharashtra Board 12th Maths Solutions Chapter 1

#### Ex.1. Write the negations of the following stating the rules used.

i) (p ∨ q) ∧ (q ∨ ~ r) ii) (p → q) ∨ r
iii) p ∧ (q ∨ r) iv) (~ p ∧ q) ∨ (p ∧ ~ q)
v) (p ∧ q) → (~ p ∨ r)
Solution :
i) ~ [(p ∨ q) ∧ (q ∨ ~ r)] ≡ ~ (p ∨ q) ∨ ~ (q ∨ ~ r) [DeMorgan’s law]
≡ (~ p ∧ ~ q) ∨ (~ q ∧ r) [DeMorgan’s law]
≡ (~ q ∧ ~ p) ∨ (~ q ∧ r) [Commutative law]
≡ ~ q ∧ (~ p ∨ r) [Distributive law]

ii) ~ [(p → q) ∨ r] ≡ ~ (p → q) ∧ ~ r [DeMorgan’s law]
≡ (p ∧ ~ q) ∧ ~ r [~ (p → q) ≡ p ∧ ~ q]

iii) ~ [p ∧ (q ∨ r)] ≡ ~ p ∨ ~ (q ∨ r) [DeMorgan’s law]
≡ ~ p ∨ (~ q ∧ ~ r) [DeMorgan’s law]

iv) ~ [(~ p ∧ q) ∨ (p ∧ ~ q)] ≡ ~ (~ p ∧ q) ∧ ~ (p ∧ ~ q) [DeMorgan’s law]
≡ (p ∨ ~ q) ∧ (~ p ∨ q) [DeMorgan’s law]

v) ~ [(p ∧ q) → (~ p ∨ r)] ≡ (p ∧ q) ∧ ~ (~ p ∨ r) [~ (p → q) ≡ p ∧ ~ q]
≡ (p ∧ q) ∧ [p ∧ ~ r] [DeMorgan’s law]
≡ q ∧ p ∧ p ∧ ~ r [Associative law
≡ q ∧ p ∧ ~ r [Idempotent law]

#### Ex.2.Rewrite the following statements without using if ...... then.

i) If prices increase then the wages rise.
ii) If it is cold, then we wear woolen clothes.
Solution :
i) Let p : Prices increase
q : The wages rise.
The given statement is p → q
but p → q ≡ ~ p ∨ q
The given statement can be written as
'Prices do not increase or the wages rise'.

ii) Let p : It is cold, q : We wear woollen clothes.
The given statement is p → q
but p → q ≡ ~ p ∨ q
The given statement can be written as
It is not cold or we wear woollen clothes.

#### Ex.3.Without using truth table prove that :

i) p ↔ q ≡ ~ (p ∧ ~ q) ∧ ~ (q ∧ ~ p)
ii) ~ (p ∨ q) ∨ (~ p ∧ q) ≡ ~ p
iii) ~ p ∧ q ≡ (p ∨ q) ∧ ~ p

Solution :
i) We know that
p ↔ q ≡ (p → q) ∧ (q → p)
≡ (~ p ∨ q) ∧ (~ q ∨ p) [Conditional law]
≡ ~ ( p ∧ ~ q) ∧ ~ (q ∧ ~p) [Demorgan’s law]

ii) ~ (p ∨ q) ∨ (~ p ∧ q) ≡ (~ p ∧ ~ q) ∨ (~ p ∧ q) [Demorgan’s law]
≡ ~ p ∧ (~ q ∨ q) [Distributive law]
≡ ~ p ∧ T [Complement law]
≡ ~ p [Identity law]

iii) (p ∧ q) ∧ ~ p ≡ ~ p ∧ (p ∨ q) [Commutative law]
≡ (~ p ∧ p) ∨ (~ p ∧ q) [Distributive law]
≡ F ∨ (~ p ∧ q) [Complement law]
≡ ~ p ∧ q [Identity law]

### Mathematical Logic - Ex 1.4 | Maharashtra Board 12th Maths Solutions Chapter 1

• mathematical logic exercise 1.4
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• mathematical logic exercise 1.2
• mathematical logic exercise 1.4
• mathematical logic exercise 1.2 solutions
• mathematical logic exercise 1.5
• mathematical logic class 12 exercise 1.4 solutions
• mathematical logic class 12 exercise 1.4 solutions