### 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
- mathematical logic exercise 1.3
- 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