Consider :
Statement I:
`(p^^~q)^^(~p^^q)` is a fallacy
Statement II: `(ptoq)harr(~q to ~p) ` 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 problem, we need to analyze the two statements given:

**Statement I:** \((p \land \neg q) \land (\neg p \land q)\) is a fallacy.

**Statement II:** \((p \to q) \iff (\neg q \to \neg p)\) is a tautology.

### Step 1: Analyze Statement I

We will create a truth table for the expression \((p \land \neg q) \land (\neg p \land q)\).

1. **List all possible truth values for \(p\) and \(q\):**
- \(p = T, q = T\)
- \(p = T, q = F\)
- \(p = F, q = T\)
- \(p = F, q = F\)

2. **Calculate \(\neg p\) and \(\neg q\):**
- If \(p = T\), then \(\neg p = F\).
- If \(p = F\), then \(\neg p = T\).
- If \(q = T\), then \(\neg q = F\).
- If \(q = F\), then \(\neg q = T\).

3. **Calculate \(p \land \neg q\) and \(\neg p \land q\):**
- For \(p = T, q = T\): \(p \land \neg q = T \land F = F\) and \(\neg p \land q = F \land T = F\).
- For \(p = T, q = F\): \(p \land \neg q = T \land T = T\) and \(\neg p \land q = F \land F = F\).
- For \(p = F, q = T\): \(p \land \neg q = F \land F = F\) and \(\neg p \land q = T \land T = T\).
- For \(p = F, q = F\): \(p \land \neg q = F \land T = F\) and \(\neg p \land q = T \land F = F\).

4. **Calculate the final expression \((p \land \neg q) \land (\neg p \land q)\):**
- For \(p = T, q = T\): \(F \land F = F\).
- For \(p = T, q = F\): \(T \land F = F\).
- For \(p = F, q = T\): \(F \land T = F\).
- For \(p = F, q = F\): \(F \land F = F\).

The final column shows that the expression is always false. Therefore, Statement I is a fallacy.

### Step 2: Analyze Statement II

Now, we will create a truth table for the expression \((p \to q) \iff (\neg q \to \neg p)\).

1. **Calculate \(p \to q\) and \(\neg q \to \neg p\):**
- \(p \to q\) is false only when \(p = T\) and \(q = F\).
- \(\neg q \to \neg p\) is false only when \(\neg q = T\) (i.e., \(q = F\)) and \(\neg p = F\) (i.e., \(p = T\)).

2. **Construct the truth table:**
- For \(p = T, q = T\): \(p \to q = T\) and \(\neg q \to \neg p = F\) → \(T \iff F = F\).
- For \(p = T, q = F\): \(p \to q = F\) and \(\neg q \to \neg p = T\) → \(F \iff T = F\).
- For \(p = F, q = T\): \(p \to q = T\) and \(\neg q \to \neg p = T\) → \(T \iff T = T\).
- For \(p = F, q = F\): \(p \to q = T\) and \(\neg q \to \neg p = T\) → \(T \iff T = T\).

The final column shows that the expression is true in all cases except when \(p = T\) and \(q = T\). Therefore, Statement II is a tautology.

### Conclusion

- Statement I is a fallacy (always false).
- Statement II is a tautology (always true).

### Final Answer

- Statement I is true.
- Statement II is true.