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: Create the truth table structure
We need to list all possible truth values for the variables \( p \) and \( q \). Since both \( p \) and \( q \) can be either true (T) or false (F), we will have four combinations of truth values.

### Step 2: Fill in the truth values for \( p \) and \( q \)
We will create the first two columns of our truth table for \( p \) and \( q \):

| \( p \) | \( q \) |
|---------|---------|
| T | T |
| T | F |
| F | T |
| F | F |

### Step 3: Calculate \( p \land q \)
Next, we will compute the conjunction \( p \land q \) (p AND q):

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

### Step 4: Calculate \( \neg (p \land q) \)
Now, we will take the negation of \( p \land q \):

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

### Step 5: Calculate \( p \lor \neg (p \land q) \)
Finally, we will compute \( p \lor \neg (p \land q) \):

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

### Conclusion
In the last column, we see that \( p \lor \neg (p \land q) \) is true (T) for all possible combinations of truth values for \( p \) and \( q \). Therefore, we conclude that the expression \( p \lor \neg (p \land q) \) is a tautology.

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