Show that `(p^^q)vv(~p)vv(p^^~q)` is a tautology

Explanation

To show that the expression \((p \land q) \lor (\neg p) \lor (p \land \neg q)\) is a tautology, we will create a truth table and evaluate the expression for all possible truth values of \(p\) and \(q\).

### Step 1: Identify the variables
The variables in the expression are \(p\) and \(q\).

### Step 2: Create a truth table
We will create a truth table that includes all possible combinations of truth values for \(p\) and \(q\). There are 2 variables, so we will have \(2^2 = 4\) rows.

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

### Step 3: Fill in the truth table
1. **Negation of \(p\) (\(\neg p\))**: If \(p\) is True (T), then \(\neg p\) is False (F), and vice versa.
2. **Negation of \(q\) (\(\neg q\))**: If \(q\) is True (T), then \(\neg q\) is False (F), and vice versa.
3. **Conjunction \(p \land q\)**: This is True only when both \(p\) and \(q\) are True.
4. **Conjunction \(p \land \neg q\)**: This is True when \(p\) is True and \(q\) is False.
5. **Final Expression**: The final expression \((p \land q) \lor (\neg p) \lor (p \land \neg q)\) is True if at least one of the components is True.

### Step 4: Analyze the final column
From the truth table, we see that the final column, which represents the expression \((p \land q) \lor (\neg p) \lor (p \land \neg q)\), is True for all combinations of \(p\) and \(q\). Therefore, the expression is a tautology.

### Conclusion
Since the expression evaluates to True for all possible truth values of \(p\) and \(q\), we conclude that \((p \land q) \lor (\neg p) \lor (p \land \neg q)\) is indeed a tautology.