Solved Practice Exercises: Construct a truth table for each ... - Free Printable

Problem Analysis:

We are tasked with constructing truth tables for two logical formulas and determining whether they are tautologies. A tautology is a formula that evaluates to true for all possible truth assignments of its variables.

The formulas are:
1. \( \neg p \lor q \to p \)
2. \( (p \to q) \lor (p \to \neg q) \)

Let's solve each part step by step.

---

Formula 1: \( \neg p \lor q \to p \)

#### Step 1: Truth Table Setup
We need to evaluate the formula \( \neg p \lor q \to p \). The truth table will have columns for:
- \( p \)
- \( q \)
- \( \neg p \)
- \( \neg p \lor q \)
- \( \neg p \lor q \to p \)

#### Step 2: Fill in the Truth Table
We evaluate each column step by step for all combinations of \( p \) and \( q \).

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

#### Explanation of Each Row:
1. Row 1: \( p = T, q = T \)
- \( \neg p = \neg T = F \)
- \( \neg p \lor q = F \lor T = T \)
- \( \neg p \lor q \to p = T \to T = T \)

2. Row 2: \( p = T, q = F \)
- \( \neg p = \neg T = F \)
- \( \neg p \lor q = F \lor F = F \)
- \( \neg p \lor q \to p = F \to T = T \)

3. Row 3: \( p = F, q = T \)
- \( \neg p = \neg F = T \)
- \( \neg p \lor q = T \lor T = T \)
- \( \neg p \lor q \to p = T \to F = F \)

4. Row 4: \( p = F, q = F \)
- \( \neg p = \neg F = T \)
- \( \neg p \lor q = T \lor F = T \)
- \( \neg p \lor q \to p = T \to F = F \)

#### Conclusion for Formula 1:
The formula \( \neg p \lor q \to p \) is not a tautology because it evaluates to false in rows 3 and 4.

---

Formula 2: \( (p \to q) \lor (p \to \neg q) \)

#### Step 1: Truth Table Setup
We need to evaluate the formula \( (p \to q) \lor (p \to \neg q) \). The truth table will have columns for:
- \( p \)
- \( q \)
- \( \neg q \)
- \( p \to q \)
- \( p \to \neg q \)
- \( (p \to q) \lor (p \to \neg q) \)

#### Step 2: Fill in the Truth Table
We evaluate each column step by step for all combinations of \( p \) and \( q \).

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

#### Explanation of Each Row:
1. Row 1: \( p = T, q = T \)
- \( \neg q = \neg T = F \)
- \( p \to q = T \to T = T \)
- \( p \to \neg q = T \to F = F \)
- \( (p \to q) \lor (p \to \neg q) = T \lor F = T \)

2. Row 2: \( p = T, q = F \)
- \( \neg q = \neg F = T \)
- \( p \to q = T \to F = F \)
- \( p \to \neg q = T \to T = T \)
- \( (p \to q) \lor (p \to \neg q) = F \lor T = T \)

3. Row 3: \( p = F, q = T \)
- \( \neg q = \neg T = F \)
- \( p \to q = F \to T = T \)
- \( p \to \neg q = F \to F = T \)
- \( (p \to q) \lor (p \to \neg q) = T \lor T = T \)

4. Row 4: \( p = F, q = F \)
- \( \neg q = \neg F = T \)
- \( p \to q = F \to F = T \)
- \( p \to \neg q = F \to T = T \)
- \( (p \to q) \lor (p \to \neg q) = T \lor T = T \)

#### Conclusion for Formula 2:
The formula \( (p \to q) \lor (p \to \neg q) \) is a tautology because it evaluates to true for all combinations of \( p \) and \( q \).

---

Final Answer:

1. \( \neg p \lor q \to p \) is not a tautology.
2. \( (p \to q) \lor (p \to \neg q) \) is a tautology.

\[
\boxed{(p \to q) \lor (p \to \neg q)}
\]