Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 1 Mathematical Logic Ex 1.2 Questions and Answers.

## Maharashtra State Board 12th Maths Solutions Chapter 1 Mathematical Logic Ex 1.2

Question 1.

Construct the truth table for each of the following statement patterns:

(i) [(p → q) ∧ q] → p

Solution :

Here are two statements and three connectives.

∴ there are 2 × 2 = 4 rows and 2 + 3 = 5 columns in the truth table.

(ii) (p ∧ ~q) ↔ (p → q)

Solution:

(iii) (p ∧ q) ↔ (q ∨ r)

Solution:

(iv) p → [~(q ∧ r)]

Solution:

(v) ~p ∧ [(p ∨ ~q ) ∧ q]

Solution:

(vi) (~p → ~q) ∧ (~q → ~p)

Solution:

(vii) (q → p) ∨ (~p ↔ q)

Solution:

(viii) [p → (q → r)] ↔ [(p ∧ q) → r]

Solution:

(ix) p → [~(q ∧ r)]

Solution:

(x) (p ∨ ~q) → (r ∧ p)

Solution:

Question 2.

Using truth tables prove the following logical equivalences.

(i) ~p ∧ q ≡ (p ∨ q) ∧ ~p

Solution:

The entries in the columns 4 and 6 are identical.

∴ ~p ∧ q ≡ (p ∨ q) ∧ ~p.

(ii) ~(p ∨ q) ∨ (~p ∧ q) ≡ ~p

Solution:

The entries in the columns 3 and 7 are identical.

∴ ~(p ∨ q) ∧ (~p ∧ q) = ~p.

(iii) p ↔ q ≡ ~[(p ∨ q) ∧ ~(p ∧ q)]

Solution:

The entries in the columns 3 and 8 are identical.

∴ p ↔ q ≡ ~[(p ∨ q) ∧ ~(p ∧ q)].

(iv) p → (q → p) ≡ ~p → (p → q)

Solution:

The entries in the columns 4 and 7 are identical.

∴ p → (q → p) ≡ ~p → (p → q).

(v) (p ∨ q ) → r ≡ (p → r) ∧ (q → r)

Solution:

The entries in the columns 5 and 8 are identical.

∴ (p ∨ q ) → r ≡ (p → r) ∧ (q → r).

(vi) p → (q ∧ r) ≡ (p → q) ∧ (p → r)

Solution:

The entries in the columns 5 and 8 are identical.

∴ p → (q ∧ r) ≡ (p → q) ∧ (p → r).

(vii) p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r)

Solution:

The entries in the columns 5 and 8 are identical.

∴ p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r).

(viii) [~(p ∨ q) ∨ (p ∨ q)] ∧ r ≡ r

Solution:

The entries in the columns 3 and 7 are identical.

∴ [~(p ∨ q) ∨ (p ∨ q)] ∧ r ≡ r.

(ix) ~(p ↔ q) ≡ (p ∧ ~q) ∨ (q ∧ ~p)

Solution:

The entries in the columns 6 and 9 are identical.

∴ ~(p ↔ q) ≡ (p ∧ ~q) ∨ (q ∧ ~p).

Question 3.

Examine whether each of the following statement patterns is a tautology or a contradiction or a contingency.

(i) (p ∧ q) → (q ∨ p)

Solution:

All the entries in the last column of the above truth table are T.

∴ (p ∧ q) → (q ∨ p) is a tautology.

(ii) (p → q) ↔ (~p ∨ q)

Solution:

All the entries in the last column of the above truth table are T.

∴ (p → q) ↔ (~p ∨ q) p is a tautology.

(iii) [~(~p ∧ ~q)] ∨ q

Solution:

The entries in the last column of the above truth table are neither all T nor all F.

∴ [~(~p ∧ ~q)] ∨ q is a contingency.

(iv) [(p → q) ∧ q)] → p

Solution:

The entries in the last column of the above truth table are neither all T nor all F.

∴ [(p → q) ∧ q)] → p is a contingency

(v) [(p → q) ∧ ~q] → ~p

Solution:

All the entries in the last column of the above truth table are T.

∴ [(p → q) ∧ ~q] → ~p is a tautology.

(vi) (p ↔ q) ∧ (p → ~q)

Solution:

The entries in the last column of the above truth table are neither all T nor all F.

∴ (p ↔ q) ∧ (p → ~q) is a contingency.

(vii) ~(~q ∧ p) ∧ q

Solution:

The entries in the last column of the above truth table are neither all T nor all F.

∴ ~(~q ∧ p) ∧ q is a contingency.

(viii) (p ∧ ~q) ↔ (p → q)

Solution:

All the entries in the last column of the above truth table are F.

∴ (p ∧ ~q) ↔ (p → q) is a contradiction.

(ix) (~p → q) ∧ (p ∧ r)

Solution:

The entries in the last column of the above truth table are neither all T nor all F.

∴ (~p → q) ∧ (p ∧ r) is a contingency.

(x) [p → (~q ∨ r)] ↔ ~[p → (q → r)]

Solution:

All the entries in the last column of the above truth table are F.

∴ [p → (~q ∨ r)] ↔ ~[p → (q → r)] is a contradiction