Balbharati Maharashtra State Board 12th Commerce Maths Solution Book Pdf Chapter 1 Mathematical Logic Ex 1.6 Questions and Answers.

## Maharashtra State Board 12th Commerce Maths Solutions Chapter 1 Mathematical Logic Ex 1.6

Question 1.

Prepare the truth tables for the following statement patterns:

(i) p → (~p ∨ q)

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 ∧ r) → (p ∨ ~q)

Solution:

Here are three statements and 4 connectives.

∴ there are 2 × 2 × 2 = 8 rows and 3 + 4 = 7 columns in the truth table.

(iv) (p ∧ q) ∨ ~r

Solution:

Question 2.

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

(i) q ∨ [~(p ∧ q)]

Solution:

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

∴ q ∨ [~(p ∧ q)] is a tautology.

(ii) (~q ∧ p) ∧ (p ∧ ~p)

Solution:

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

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

(iii) (p ∧ ~q) → (~p ∧ ~q)

Solution:

The entries in the last column are neither all T nor all F.

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

(iv) ~p → (p → ~q)

Solution:

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

∴ p → (p → ~q) is a tautology.

Question 3.

Prove that each of the following statement pattern is a tautology:

(i) (p ∧ q) → q

Solution:

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

∴ (p ∧ q) → q is a tautology.

(ii) (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.

(iii) (~p ∧ ~q) → (p → q)

Solution:

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

∴ (~p ∧ ~q) → (p → q) is a tautology.

(iv) (~p ∨ ~q) ↔ ~(p ∧ q)

Solution:

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

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

Question 4.

Prove that each of the following statement pattern is a contradiction:

(i) (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.

(ii) (p ∧ q) ∧ ~p

Solution:

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

∴ (p ∧ q) ∧ ~p is a contradiction.

(iii) (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.

(iv) (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.

Question 5.

Show that each of the following statement pattern is a contingency:

(i) (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.

(ii) (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.

(iii) p ∧ [(p → ~q) → q]

Solution:

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

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

(iv) (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.

Question 6.

Using the truth table, verify:

(i) p ∨ (q ∧ r) = (p ∨ q) ∧ (p ∨ r)

Solution:

The entries in columns 5 and 8 are identical.

∴ p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r).

(ii) p → (p → q) ≡ ~q → (p → q)

Solution:

The entries in columns 5 and 6 are identical.

∴ p → (p → q) ≡ ~q → (p → q)

(iii) ~(p → ~q) ≡ p ∧ ~(~q) ≡ p ∧ q

Solution:

The entries in columns 5, 7 and 8 are identical.

∴ ~(p → ~q) ≡ p ∧ ~(~q) ≡ p ∧ q.

(iv) ~(p ∨ q) ∨ (~p ∧ q) ≡ ~p

Solution:

The entries in columns 3 and 7 are identical.

∴ ~(p ∨ q) ∨ (~p ∧ q) ≡ ~p.

Question 7.

Prove that the following pairs of statement patterns are equivalent:

(i) p ∨ (q ∧ r) and (p ∨ q) ∧ (p ∨ r)

Solution:

The entries in columns 5 and 8 are identical.

∴ p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r)

(ii) p ↔ q and (p → q) ∧ (q → p)

Solution:

The entries in columns 3 and 6 are identical.

∴ p ↔ q ≡ (p → q) ∧ (q → p)

(iii) p → q and ~q → ~p and ~p ∨ q

Solution:

The entries in columns 5, 6 and 7 are identical.

∴ p → q ≡ ~q → ~p ≡ ~p ∨ q.

(iv) ~(p ∧ q) and ~p ∨ ~q

Solution:

The entries in columns 6 and 7 are identical.

∴ ~(p ∧ q) ≡ ~p ∨ ~q.