If p and q are two statements, then which of the following statement is a tautology

Options

A. `p rArr (p vv ~q)`
B. `(p vv q) rArr p`
C. `p rArr (p^^q)`
D. `phArr(prArrq)`

Answer

To determine which of the given statements is a tautology, we will analyze each option using truth tables. A tautology is a statement that is always true, regardless of the truth values of its components.

### Step-by-Step Solution

1. **Identify the Statements**: We have two statements \( p \) and \( q \). We need to evaluate the given options to find which one is a tautology.

2. **Construct the Truth Table**:
- The truth values for \( p \) and \( q \) can be represented as follows:
- \( p \): True (T), True (T), False (F), False (F)
- \( q \): True (T), False (F), True (T), False (F)

3. **Negation of \( q \)**:
- The negation of \( q \) (denoted as \( \neg q \)) will be:
- \( \neg q \): False (F), True (T), False (F), True (T)

4. **Evaluate Each Option**:
- **Option 1**: \( p \implies (p \lor \neg q) \)
- Construct the truth table:
- For \( p = T, q = T \): \( p \lor \neg q = T \) → \( T \implies T = T \)
- For \( p = T, q = F \): \( p \lor \neg q = T \) → \( T \implies T = T \)
- For \( p = F, q = T \): \( p \lor \neg q = F \) → \( F \implies F = T \)
- For \( p = F, q = F \): \( p \lor \neg q = T \) → \( F \implies T = T \)
- Result: All cases are true, so this option is a tautology.

- **Option 2**: \( (p \lor q) \implies p \)
- Construct the truth table:
- For \( p = T, q = T \): \( p \lor q = T \) → \( T \implies T = T \)
- For \( p = T, q = F \): \( p \lor q = T \) → \( T \implies T = T \)
- For \( p = F, q = T \): \( p \lor q = T \) → \( T \implies F = F \)
- For \( p = F, q = F \): \( p \lor q = F \) → \( F \implies F = T \)
- Result: Not all cases are true, so this option is not a tautology.

- **Option 3**: \( p \implies (p \land q) \)
- Construct the truth table:
- For \( p = T, q = T \): \( p \land q = T \) → \( T \implies T = T \)
- For \( p = T, q = F \): \( p \land q = F \) → \( T \implies F = F \)
- For \( p = F, q = T \): \( p \land q = F \) → \( F \implies F = T \)
- For \( p = F, q = F \): \( p \land q = F \) → \( F \implies F = T \)
- Result: Not all cases are true, so this option is not a tautology.

- **Option 4**: \( p \iff (p \implies q) \)
- Construct the truth table:
- For \( p = T, q = T \): \( p \implies q = T \) → \( T \iff T = T \)
- For \( p = T, q = F \): \( p \implies q = F \) → \( T \iff F = F \)
- For \( p = F, q = T \): \( p \implies q = T \) → \( F \iff T = F \)
- For \( p = F, q = F \): \( p \implies q = T \) → \( F \iff T = F \)
- Result: Not all cases are true, so this option is not a tautology.

5. **Conclusion**: The only option that is a tautology is **Option 1**: \( p \implies (p \lor \neg q) \).