The statement `(p implies p) ^^ (~p implies p)` is a-

Options

A. Tautology and contradiction
B. Neither tautology nor contradication
C. Contradiction
D. Tautology

Answer

To determine whether the statement `(p implies p) ^^ (~p implies p)` is a tautology, contradiction, or neither, we will construct a truth table.

### Step 1: Define the components of the statement
The statement consists of two parts:
1. \( p \implies p \)
2. \( \neg p \implies p \)

### Step 2: Create a truth table
We will create a truth table for \( p \), \( \neg p \), \( p \implies p \), and \( \neg p \implies p \).

| p | ¬p | p ⇒ p | ¬p ⇒ p |
|-------|-------|-------|--------|
| T | F | T | T |
| F | T | T | F |

### Step 3: Calculate \( p \implies p \)
The implication \( p \implies p \) is always true regardless of the truth value of \( p \):
- When \( p \) is True (T), \( p \implies p \) is True (T).
- When \( p \) is False (F), \( p \implies p \) is still True (T).

### Step 4: Calculate \( \neg p \implies p \)
Now, we calculate \( \neg p \implies p \):
- When \( p \) is True (T), \( \neg p \) is False (F), so \( \neg p \implies p \) is True (T).
- When \( p \) is False (F), \( \neg p \) is True (T), so \( \neg p \implies p \) is False (F).

### Step 5: Combine the results using AND (^^)
Now we combine the results of \( p \implies p \) and \( \neg p \implies p \) using AND (^^):

| p | ¬p | p ⇒ p | ¬p ⇒ p | (p ⇒ p) ^^ (¬p ⇒ p) |
|-------|-------|-------|--------|---------------------|
| T | F | T | T | T |
| F | T | T | F | F |

### Step 6: Analyze the final column
The final column shows the results of the combined statement:
- When \( p \) is True, the result is True.
- When \( p \) is False, the result is False.

Since the combined statement is not always true (it is false when \( p \) is False), it is neither a tautology (which would require all true values) nor a contradiction (which would require all false values).

### Conclusion
Thus, the statement `(p implies p) ^^ (~p implies p)` is neither a tautology nor a contradiction.