**Chapter 1 Mathematical Logic Ex 1.4**

## Chapter 1 Mathematical Logic Ex 1.4

**Question 1.Using rules of negation write the negations of the following with justification.(i) ~q → pSolution:**

The negation of ~q → p is

~(~q → p) ≡ ~ q ∧ ~p…. (Negation of implication)

**(ii) p ∧ ~qSolution:**

The negation of p ∧ ~q is

~(p ∧ ~q) ≡ ~p ∨ ~(~q) … (Negation of conjunction)

≡ ~ p ∨ q … (Negation of negation)

**(iii) p ∨ ~qSolution:**

The negation of p ∨ ~ p is

~ (p ∨ ~(q) ≡ ~p ∧ ~(~(q) … (Negation of disjunction)

≡ ~ p ∧ q … (Negation of negation)

**(iv) (p ∨ ~q) ∧ rSolution:**

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) ≡ pSolution:**
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 ∨ qSolution:**
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.