Prove by construction of truth table that `p vv ~ (p ^^q)` is a tautology

Answer

To prove that the expression \( p \lor \neg (p \land q) \) is a tautology using a truth table, we will follow these steps:

### Step 1: Set up the truth table
We will create a truth table with columns for \( p \), \( q \), \( \neg p \), \( p \land q \), \( \neg (p \land q) \), and \( p \lor \neg (p \land q) \).

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

### Step 2: Fill in the truth values
1. **Column for \( p \)**: List all combinations of truth values for \( p \) (True or False).
2. **Column for \( q \)**: List all combinations of truth values for \( q \).
3. **Column for \( \neg p \)**: This is the negation of \( p \). If \( p \) is True, \( \neg p \) is False, and vice versa.
4. **Column for \( p \land q \)**: This is True only when both \( p \) and \( q \) are True.
5. **Column for \( \neg (p \land q) \)**: This is the negation of the previous column. It is True when \( p \land q \) is False.
6. **Column for \( p \lor \neg (p \land q) \)**: This is True if either \( p \) is True or \( \neg (p \land q) \) is True.

### Step 3: Analyze the final column
In the final column \( p \lor \neg (p \land q) \), we see that it is True for all combinations of truth values for \( p \) and \( q \). Since the expression evaluates to True in every possible case, we conclude that \( p \lor \neg (p \land q) \) is a tautology.

### Conclusion
Thus, we have proven that \( p \lor \neg (p \land q) \) is a tautology by constructing a truth table.

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