Which of the following is a tautology?

Options

A. `(p rarr q) ^^ ( q rarr p)`
B. `(~p) ^^(p vv q)rarr q`
C. `(q rarr p) vv ~ ( p rarr q)`
D. `(~q) vv ( p ^^ q) rarr q`

Answer

To determine which of the given options is a tautology, we need to 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:

**Step 1: Analyze Option 1: \( p \implies (q \cap (q \implies p)) \)**

1. Create a truth table for \( p \) and \( q \):
- \( p \): True, True, False, False
- \( q \): True, False, True, False

2. Calculate \( q \implies p \):
- True (T) implies True (T) = T
- False (F) implies True (T) = T
- True (T) implies False (F) = F
- False (F) implies False (F) = T

3. Calculate \( q \cap (q \implies p) \):
- T and T = T
- F and T = F
- T and F = F
- F and T = F

4. Calculate \( p \implies (q \cap (q \implies p)) \):
- T implies T = T
- T implies F = F
- F implies F = T
- F implies F = T

5. Result: The truth values are T, F, T, T. Since there are false values, this is **not a tautology**.

**Step 2: Analyze Option 2: \( \neg p \cup q \)**

1. Create a truth table for \( p \) and \( q \):
- \( p \): True, True, False, False
- \( q \): True, False, True, False

2. Calculate \( \neg p \):
- F, F, T, T

3. Calculate \( \neg p \cup q \):
- F or T = T
- F or F = F
- T or T = T
- T or F = T

4. Result: The truth values are T, F, T, T. Since there are false values, this is **not a tautology**.

**Step 3: Analyze Option 3: \( p \implies q \) and \( q \implies p \)**

1. Create a truth table for \( p \) and \( q \):
- \( p \): True, True, False, False
- \( q \): True, False, True, False

2. Calculate \( p \implies q \):
- T implies T = T
- T implies F = F
- F implies T = T
- F implies F = T

3. Calculate \( q \implies p \):
- T implies T = T
- F implies T = T
- T implies F = F
- F implies F = T

4. Result: The truth values are T, F, T, T. Since there are false values, this is **not a tautology**.

**Step 4: Analyze Option 4: \( p \cap \neg q \)**

1. Create a truth table for \( p \) and \( q \):
- \( p \): True, True, False, False
- \( q \): True, False, True, False

2. Calculate \( \neg q \):
- F, T, F, T

3. Calculate \( p \cap \neg q \):
- T and F = F
- T and T = T
- F and F = F
- F and T = F

4. Result: The truth values are F, T, F, F. Since there are false values, this is **not a tautology**.

### Conclusion:
After analyzing all options, none of them are tautologies.