Statements -1 : `~(pharr~q)` is equivalent to `p harr q`
Statement-2: `~( p harr ~q)` is a tautology.

Options

A. Statement-1 is true, statement 2 is true, statement 2 is a correct explanation for statement 1
B. Statement 1 is true, statement-2 is true, statement 2 is not a correct explanation for statement 1
C. Statement 1 is true , statement 2 is false,
D. statement 1 is false, statement 2 is true

Explanation

To solve the question, we will analyze the two statements using truth tables.

### Step 1: Define the Statements
1. Statement 1: `~(p ↔ ~q)` is equivalent to `p ↔ q`.
2. Statement 2: `~(p ↔ ~q)` is a tautology.

### Step 2: Create a Truth Table
We need to create a truth table for the variables `p` and `q`, and evaluate the expressions `~(p ↔ ~q)` and `p ↔ q`.

| p | q | ~q | p ↔ ~q | ~(p ↔ ~q) | p ↔ q |
|-------|-------|-------|--------|------------|--------|
| T | T | F | F | T | T |
| T | F | T | T | F | F |
| F | T | F | T | F | F |
| F | F | T | F | T | T |

### Step 3: Analyze the Truth Table
- **Column for `~(p ↔ ~q)`**:
- The values are T, F, F, T.
- **Column for `p ↔ q`**:
- The values are T, F, F, T.

### Step 4: Check Equivalence for Statement 1
From the truth table:
- The values for `~(p ↔ ~q)` and `p ↔ q` are the same (T, F, F, T).
- Therefore, Statement 1 is **true**: `~(p ↔ ~q)` is equivalent to `p ↔ q`.

### Step 5: Check for Tautology for Statement 2
A tautology is a statement that is always true regardless of the truth values of its components.
- The values for `~(p ↔ ~q)` are T, F, F, T.
- Since not all values are true, Statement 2 is **false**: `~(p ↔ ~q)` is not a tautology.

### Conclusion
- Statement 1 is correct.
- Statement 2 is incorrect.

### Final Answer
Thus, the answer is that Statement 1 is true and Statement 2 is false.

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