The number of choices for `Deltain{^^,vv, implies,hArr}` , such that `(p Deltaq) = ((p Delta ~q) vv ((~p) Delta q))` is a tautology, is :

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the problem, we need to determine the number of choices for the operator \( \Delta \) from the set \( \{ \land, \lor, \implies, \hArr \} \) such that the expression \( (p \Delta q) = (p \Delta \neg q) \lor (\neg p \Delta q) \) is a tautology.

### Step-by-Step Solution:

1. **Understanding the Expression**:
We need to analyze the expression \( (p \Delta q) = (p \Delta \neg q) \lor (\neg p \Delta q) \). This means that for all truth values of \( p \) and \( q \), the left-hand side (LHS) must equal the right-hand side (RHS).

2. **Substituting Operators**:
We will substitute each operator from the set \( \{ \land, \lor, \implies, \hArr \} \) into \( \Delta \) and check if the equality holds for all truth values of \( p \) and \( q \).

3. **Testing Each Operator**:
- **For \( \Delta = \land \)**:
\[ (p \land q) = (p \land \neg q) \lor (\neg p \land q) \]
This does not hold true for all combinations of truth values.

- **For \( \Delta = \lor \)**:
\[ (p \lor q) = (p \lor \neg q) \lor (\neg p \lor q) \]
This does not hold true for all combinations of truth values.

- **For \( \Delta = \implies \)**:
\[ (p \implies q) = (p \implies \neg q) \lor (\neg p \implies q) \]
This does hold true for all combinations of truth values. (True when \( p \) is false or \( q \) is true.)

- **For \( \Delta = \hArr \)**:
\[ (p \hArr q) = (p \hArr \neg q) \lor (\neg p \hArr q) \]
This does not hold true for all combinations of truth values.

4. **Conclusion**:
The only operator \( \Delta \) that makes the expression a tautology is \( \implies \).

### Final Answer:
The number of choices for \( \Delta \) such that \( (p \Delta q) = (p \Delta \neg q) \lor (\neg p \Delta q) \) is a tautology is **1** (only \( \implies \)).