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

Options

A. a contradiction
B. a tautology
C. either (1) or (2)
D. neither (1) nor (2)

Explanation

To determine whether the expression \((p \land \neg q) \land (\neg p \land q)\) is a contradiction, tautology, or neither, we can construct a truth table. Here's a step-by-step solution:

### Step 1: Identify the Variables
We have two variables, \(p\) and \(q\).

### Step 2: List All Possible Truth Values
For two variables, there are four possible combinations of truth values:
1. \(p = T\), \(q = T\)
2. \(p = T\), \(q = F\)
3. \(p = F\), \(q = T\)
4. \(p = F\), \(q = F\)

### Step 3: Calculate Negations
We need to find \(\neg p\) 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: Construct the Truth Table
Now we will create the truth table including \(p\), \(q\), \(\neg p\), \(\neg q\), \(p \land \neg q\), \(\neg p \land q\), and finally \((p \land \neg q) \land (\neg p \land q)\).

| \(p\) | \(q\) | \(\neg p\) | \(\neg q\) | \(p \land \neg q\) | \(\neg p \land q\) | \((p \land \neg q) \land (\neg p \land q)\) |
|-------|-------|------------|------------|---------------------|---------------------|----------------------------------------------|
| 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 5: Analyze the Final Column
The final column represents the expression \((p \land \neg q) \land (\neg p \land q)\). We see that it is false for all combinations of truth values.

### Conclusion
Since the expression is false for all possible truth values, it is a contradiction.

### Final Answer
The expression \((p \land \neg q) \land (\neg p \land q)\) is a contradiction.

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