The proposition ` p to ~ (p^^~ q)` is

Options

A. a contradiction
B. a tautology
C. either (1) or (2)
D. neither (1) nor (2)

Explanation

To determine the nature of the proposition \( p \to \sim (p \land \sim q) \), we will analyze it step by step using a truth table.

### Step-by-Step Solution:

1. **Identify the Variables**:
We have two variables: \( p \) and \( q \).

2. **Determine the Number of Possible Outcomes**:
Since there are two variables, the number of possible outcomes is \( 2^2 = 4 \).

3. **Construct the Truth Table**:
We will create a truth table with columns for \( p \), \( q \), \( \sim q \), \( p \land \sim q \), and \( p \to \sim (p \land \sim q) \).

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

4. **Fill in the Truth Table**:
- For \( \sim q \): The negation of \( q \).
- For \( p \land \sim q \): This is true only if both \( p \) is true and \( \sim q \) is true.
- For \( \sim (p \land \sim q) \): This is the negation of \( p \land \sim q \).
- For \( p \to \sim (p \land \sim q) \): This is true unless \( p \) is true and \( \sim (p \land \sim q) \) is false.

5. **Analyze the Results**:
- The last column represents the truth values of the proposition \( p \to \sim (p \land \sim q) \).
- The results are: T, F, T, T.

6. **Determine the Nature of the Proposition**:
- Since the proposition is not always true (not a tautology) and not always false (not a contradiction), it falls into the category of "neither".

### Conclusion:
The proposition \( p \to \sim (p \land \sim q) \) is neither a tautology nor a contradiction.