Let `Delta,nabla in{^^,vv}` be such that `(p rarr q)Delta(p nabla q)` is a tautology.Then

Options

A. `Delta=^^,nabla=^^`
B. `Delta=vv,nabla=vv`
C. `Delta=vv,nabla=^^`
D. `Delta=^^,nabla=vv`

Answer

To solve the problem, we need to determine the conditions under which the expression \((p \rightarrow q) \Delta (p \nabla q)\) is a tautology. Here, \(\Delta\) and \(\nabla\) are logical operators that we need to define.

### Step-by-Step Solution:

1. **Understanding the Implication**:
The expression \(p \rightarrow q\) can be rewritten using logical equivalence:
\[ p \rightarrow q \equiv \neg p \lor q \]

2. **Defining the Operators**:
We need to define the operators \(\Delta\) and \(\nabla\). Let's assume:
- \(\Delta\) is the logical operator representing union (logical OR).
- \(\nabla\) is the logical operator representing intersection (logical AND).

3. **Rewriting the Expression**:
Substitute the definitions of \(\Delta\) and \(\nabla\) into the expression:
\[ (p \rightarrow q) \Delta (p \nabla q) \equiv (\neg p \lor q) \lor (p \land q) \]

4. **Simplifying the Expression**:
We can simplify the expression:
\[ (\neg p \lor q) \lor (p \land q) = \neg p \lor q \lor (p \land q) \]
Using the distributive property:
\[ = \neg p \lor (q \lor (p \land q)) = \neg p \lor q \]

5. **Analyzing the Tautology**:
For the expression \(\neg p \lor q\) to be a tautology, it must be true for all truth values of \(p\) and \(q\). This is only true if \(q\) is always true regardless of the value of \(p\).

6. **Conclusion**:
Therefore, the condition for \((p \rightarrow q) \Delta (p \nabla q)\) to be a tautology is that \(q\) must be true. This leads us to conclude that the operators \(\Delta\) and \(\nabla\) must be defined such that the expression holds true under all conditions.

### Final Answer:
The correct interpretation of \(\Delta\) and \(\nabla\) that makes the expression a tautology is:
- \(\Delta = \text{union (logical OR)}\)
- \(\nabla = \text{intersection (logical AND)}\)