Prove by constructing truth table that `(p ^^q) ^^ ``~ (p vv q)` is fallacy (contradiction)

Answer

To prove that the expression \((p \land q) \land \neg (p \lor q)\) is a fallacy (contradiction), we will construct a truth table. A fallacy means that the expression evaluates to false for all possible truth values of \(p\) and \(q\).

### Step 1: Set Up the Truth Table

We will create a truth table with columns for \(p\), \(q\), \(p \land q\), \(p \lor q\), \(\neg (p \lor q)\), and finally \((p \land q) \land \neg (p \lor q)\).

| \(p\) | \(q\) | \(p \land q\) | \(p \lor q\) | \(\neg (p \lor q)\) | \((p \land q) \land \neg (p \lor q)\) |
|-------|-------|----------------|---------------|----------------------|--------------------------------------|
| T | T | T | T | F | F |
| T | F | F | T | F | F |
| F | T | F | T | F | F |
| F | F | F | F | T | F |

### Step 2: Fill in the Truth Values

1. **Column for \(p\) and \(q\)**: We list all combinations of truth values for \(p\) and \(q\).
- \(T, T\)
- \(T, F\)
- \(F, T\)
- \(F, F\)

2. **Column for \(p \land q\)**: This column is true only if both \(p\) and \(q\) are true.
- \(T \land T = T\)
- \(T \land F = F\)
- \(F \land T = F\)
- \(F \land F = F\)

3. **Column for \(p \lor q\)**: This column is true if at least one of \(p\) or \(q\) is true.
- \(T \lor T = T\)
- \(T \lor F = T\)
- \(F \lor T = T\)
- \(F \lor F = F\)

4. **Column for \(\neg (p \lor q)\)**: This column is the negation of the previous column.
- \(\neg T = F\)
- \(\neg T = F\)
- \(\neg T = F\)
- \(\neg F = T\)

5. **Final Column for \((p \land q) \land \neg (p \lor q)\)**: This column combines the results of \(p \land q\) and \(\neg (p \lor q)\).
- \(T \land F = F\)
- \(F \land F = F\)
- \(F \land F = F\)
- \(F \land T = F\)

### Step 3: Conclusion

From the last column of the truth table, we see that \((p \land q) \land \neg (p \lor q)\) is false for all combinations of truth values for \(p\) and \(q\). Therefore, we conclude that the expression is indeed a fallacy (contradiction).