Consider two statements
`S_1:~p to (~q implies~p)` is a tautology
`S_2:(~q^^p) to q` is a fallacy then

Options

A. Statement I is true,statement II is false
B. Statement I is false ,statement II is true
C. Both true
D. Both false

Answer

To solve the problem, we need to analyze the two statements \( S_1 \) and \( S_2 \) given in the question.

### Statement Analysis

1. **Statement \( S_1: \neg p \to (\neg q \to \neg p) \) is a tautology.**
2. **Statement \( S_2: (\neg q \land p) \to q \) is a fallacy.**

### Step 1: Analyze Statement \( S_1 \)

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

**Construct the truth table for \( S_1 \):**

| \( p \) | \( q \) | \( \neg p \) | \( \neg q \) | \( \neg q \to \neg p \) | \( \neg p \to (\neg q \to \neg p) \) |
|---------|---------|---------------|---------------|--------------------------|-------------------------------------|
| T | T | F | F | T | T |
| T | F | F | T | F | T |
| F | T | T | F | T | T |
| F | F | T | T | T | T |

**Evaluate the last column:**
- The last column shows that \( S_1 \) evaluates to true for all combinations of truth values for \( p \) and \( q \). Therefore, \( S_1 \) is indeed a tautology.

### Step 2: Analyze Statement \( S_2 \)

Now we need to check if \( S_2 \) is a fallacy. A statement is a fallacy if it evaluates to false for all possible truth values of its variables.

**Construct the truth table for \( S_2 \):**

| \( p \) | \( q \) | \( \neg q \) | \( \neg q \land p \) | \( (\neg q \land p) \to q \) |
|---------|---------|---------------|------------------------|--------------------------------|
| T | T | F | F | T |
| T | F | T | T | F |
| F | T | F | F | T |
| F | F | T | F | T |

**Evaluate the last column:**
- The last column shows that \( S_2 \) is false only when \( p \) is true and \( q \) is false. Therefore, \( S_2 \) is not a fallacy since it does not evaluate to false for all combinations of truth values.

### Conclusion

- \( S_1 \) is a tautology (true for all cases).
- \( S_2 \) is not a fallacy (it is true in some cases).

Thus, the final answer is that both statements are false.