Statements `(p to q) harr (q to p)`

Options

A. is contradiction
B. is tautology
C. is neither contradiction not tautology
D. None of these

Explanation

To solve the statement \( (p \to q) \leftrightarrow (q \to p) \), we need to analyze the truth values of both implications and their equivalence.

### Step-by-Step Solution:

1. **Understanding the Implications**:
- The implication \( p \to q \) is true in all cases except when \( p \) is true and \( q \) is false.
- The implication \( q \to p \) is true in all cases except when \( q \) is true and \( p \) is false.

2. **Constructing the Truth Table**:
We will create a truth table for \( p \) and \( q \) to evaluate \( (p \to q) \) and \( (q \to p) \).

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

3. **Evaluating the Truth Table**:
- For \( p = T \) and \( q = T \): \( (p \to q) \) is T, \( (q \to p) \) is T, so \( (p \to q) \leftrightarrow (q \to p) \) is T.
- For \( p = T \) and \( q = F \): \( (p \to q) \) is F, \( (q \to p) \) is T, so \( (p \to q) \leftrightarrow (q \to p) \) is F.
- For \( p = F \) and \( q = T \): \( (p \to q) \) is T, \( (q \to p) \) is F, so \( (p \to q) \leftrightarrow (q \to p) \) is F.
- For \( p = F \) and \( q = F \): \( (p \to q) \) is T, \( (q \to p) \) is T, so \( (p \to q) \leftrightarrow (q \to p) \) is T.

4. **Conclusion**:
- The final column shows that the expression \( (p \to q) \leftrightarrow (q \to p) \) is true for the combinations \( (T, T) \) and \( (F, F) \), but false for \( (T, F) \) and \( (F, T) \).
- Since the expression is not always true (it is false in some cases), it is neither a tautology nor a contradiction.

### Final Answer:
The statement \( (p \to q) \leftrightarrow (q \to p) \) is **neither a tautology nor a contradiction**.