`(p^^~q)^^(~p^^q)` is

Options

A. a tautology
B. a contradiction
C. neither a tautology nor a contradiction
D. None of these

Explanation

To determine whether the expression \((p \land \neg q) \land (\neg p \land q)\) is a tautology, contradiction, or neither, we can evaluate it step by step using a truth table.

### Step 1: Define the Variables
Let:
- \(p\) be a proposition that can be either true (T) or false (F).
- \(q\) be another proposition that can also be either true (T) or false (F).

### Step 2: List All Possible Truth Values
We will consider all possible combinations of truth values for \(p\) and \(q\):
1. \(p = T\), \(q = T\)
2. \(p = T\), \(q = F\)
3. \(p = F\), \(q = T\)
4. \(p = F\), \(q = F\)

### Step 3: Calculate \(\neg p\) and \(\neg q\)
For each combination, we will find the values of \(\neg p\) and \(\neg q\):
- If \(p = T\), then \(\neg p = F\)
- If \(p = F\), then \(\neg p = T\)
- If \(q = T\), then \(\neg q = F\)
- If \(q = F\), then \(\neg q = T\)

### Step 4: Evaluate \(p \land \neg q\) and \(\neg p \land q\)
Now we will evaluate the two components of our expression:
1. \(p \land \neg q\)
2. \(\neg p \land q\)

### Step 5: Combine the Results
Finally, we will evaluate the entire expression \((p \land \neg q) \land (\neg p \land q)\).

### Truth Table
| \(p\) | \(q\) | \(\neg p\) | \(\neg q\) | \(p \land \neg q\) | \(\neg p \land q\) | \((p \land \neg q) \land (\neg p \land q)\) |
|-------|-------|------------|------------|---------------------|---------------------|-----------------------------------------------|
| T | T | F | F | F | F | F |
| T | F | F | T | T | F | F |
| F | T | T | F | F | T | F |
| F | F | T | T | F | F | F |

### Step 6: Analyze the Results
From the truth table, we can see that the final column (which represents \((p \land \neg q) \land (\neg p \land q)\)) has all values as False (F). This means that the expression is always false regardless of the truth values of \(p\) and \(q\).

### Conclusion
Since the expression \((p \land \neg q) \land (\neg p \land q)\) is false for all possible truth values of \(p\) and \(q\), it is a **contradiction**.