`{:(,"List - I",,"List - II"),((1),~(~p) harr "p is " ,(a),(~p vv ~q)),((2),(p ^^ q)^^ (~(p vv q))" is ",(b),"a tautology"),((3),~(p ^^ q)-=,(c ),(~p ^^ ~q)),(,,(d),"a contradiction"):}`
The correct match is :

Options

A. 1-b, 2-a, 3-d
B. 1-b, 2-d, 3-a
C. 1-d, 2-b, 3-a
D. 1-c, 2-a, 3-d

Answer

To solve the problem, we need to match the statements in List - I with their corresponding descriptions in List - II. Let's analyze each statement in List - I and determine its nature (tautology, contradiction, etc.) based on logical reasoning.

### Step-by-Step Solution:

1. **Statement 1: ~p ↔ p**
- This statement is a logical equivalence that states "not p if and only if p".
- This is a contradiction because there cannot be a case where both p and not p are true at the same time.
- Therefore, this matches with (d) "a contradiction".

2. **Statement 2: (p ∧ q) ∨ ~(p ∨ q)**
- This statement can be simplified using logical identities.
- The expression ~(p ∨ q) can be rewritten using De Morgan's laws as ~p ∧ ~q.
- Thus, the statement becomes (p ∧ q) ∨ (~p ∧ ~q), which is true in all cases (either both p and q are true, or both are false).
- Therefore, this matches with (b) "a tautology".

3. **Statement 3: ~(p ∧ q) ↔ (~p ∧ ~q)**
- This statement can also be analyzed using De Morgan's laws.
- The negation of (p ∧ q) is equivalent to ~p ∨ ~q, not ~p ∧ ~q.
- Therefore, this statement is false, which means it is a contradiction.
- Thus, this matches with (c) "a contradiction".

4. **Statement 4: ~(p ∨ q) ↔ (~p ∧ ~q)**
- This statement is also analyzed using De Morgan's laws.
- The negation of (p ∨ q) is equivalent to ~p ∧ ~q.
- This is true, which means it is a tautology.
- Thus, this matches with (a) "p is a tautology".

### Final Matching:
- (1) ~p ↔ p → (d) a contradiction
- (2) (p ∧ q) ∨ ~(p ∨ q) → (b) a tautology
- (3) ~(p ∧ q) ↔ (~p ∧ ~q) → (c) a contradiction
- (4) ~(p ∨ q) ↔ (~p ∧ ~q) → (a) p is a tautology

### Summary of Matches:
- (1) → (d)
- (2) → (b)
- (3) → (c)
- (4) → (a)