**Question 1.Construct the truth table for each of the following statement patterns:(i) [(p → q) ∧ q] → pSolution :**

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) ∧ ~pSolution:**

The entries in the columns 4 and 6 are identical.

∴ ~p ∧ q ≡ (p ∨ q) ∧ ~p.

**(ii) ~(p ∨ q) ∨ (~p ∧ q) ≡ ~pSolution:**

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 ≡ rSolution:**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)] ∨ qSolution:**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)] → pSolution:**

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] → ~pSolution:**

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) ∧ qSolution:**

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