(S1) `(p rArr q)vv(p^^(~q))` is a tautology
(S2) `((~ p) rArr(~q))^^((~p)vvq))` is a contradiction. Then

Options

A. both (S1) and (S2) are correct
B. Only (S2) is correct
C. Only (S1) is correct
D. both (S1) and (S2) are wrong

Answer

To solve the problem, we need to analyze the two statements given:

1. \( S_1: (p \Rightarrow q) \lor (p \land \neg q) \) is a tautology.
2. \( S_2: (\neg p \Rightarrow \neg q) \land (\neg p \lor q) \) is a contradiction.

### Step 1: Analyze Statement S1

We need to determine if \( S_1 \) is a tautology. A statement is a tautology if it evaluates to true for all possible truth values of its variables.

**Truth Table for S1:**

| \( p \) | \( q \) | \( p \Rightarrow q \) | \( \neg q \) | \( p \land \neg q \) | \( S_1 \) |
|---------|---------|-----------------------|---------------|-----------------------|-----------|
| T | T | T | F | F | T |
| T | F | F | T | T | T |
| F | T | T | F | F | T |
| F | F | T | T | F | T |

From the truth table, we see that \( S_1 \) evaluates to true for all combinations of \( p \) and \( q \). Therefore, \( S_1 \) is a tautology.

### Step 2: Analyze Statement S2

Next, we analyze \( S_2 \) to see if it is a contradiction. A statement is a contradiction if it evaluates to false for all possible truth values of its variables.

**Truth Table for S2:**

| \( p \) | \( q \) | \( \neg p \) | \( \neg q \) | \( \neg p \Rightarrow \neg q \) | \( \neg p \lor q \) | \( S_2 \) |
|---------|---------|---------------|---------------|----------------------------------|----------------------|-----------|
| T | T | F | F | T | T | T |
| T | F | F | T | T | F | F |
| F | T | T | F | F | T | F |
| F | F | T | T | T | T | T |

From the truth table, we see that \( S_2 \) evaluates to false for two combinations of \( p \) and \( q \) (when \( p \) is true and \( q \) is false, and when \( p \) is false and \( q \) is true). Therefore, \( S_2 \) is not a contradiction, as it does not evaluate to false for all combinations.

### Conclusion

- \( S_1 \) is a tautology.
- \( S_2 \) is not a contradiction.

### Final Answer

Both statements are not correct as per the initial claim. The correct conclusion is that \( S_1 \) is a tautology and \( S_2 \) is not a contradiction.