If `q` is false and `(p^^q)harr r` is also true then which of the following are tautology (A) `(pvvr)-gt(p^^r)` (B) `(pvvr)` (C) `(p^^r)-gt(pvvr)` (D) `p^^r`

Options

A. `p wedge r `
B. `p vv r`
C. `(pvvr) to (p wedge r)`
D. `(p wedge r) to (p vv r)`

Explanation

To solve the problem, we need to analyze the given statements and determine which of them is a tautology given that \( q \) is false and \( (p \land q) \iff r \) is true.

### Step-by-Step Solution:

1. **Understanding the Given Conditions:**
- We know that \( q \) is false.
- The expression \( (p \land q) \iff r \) is true.
- Since \( q \) is false, \( (p \land q) \) will also be false regardless of the value of \( p \). Therefore, for the equivalence \( (p \land q) \iff r \) to hold true, \( r \) must also be false.

2. **Evaluating Each Option:**
- **Option (A):** \( (p \lor r) \to (p \land r) \)
- Since \( r \) is false, \( (p \lor r) \) simplifies to \( p \lor \text{false} = p \).
- \( (p \land r) \) simplifies to \( (p \land \text{false}) = \text{false} \).
- Thus, we have \( p \to \text{false} \), which is only true when \( p \) is false. Therefore, this is not a tautology.

- **Option (B):** \( (p \lor r) \)
- This simplifies to \( p \lor \text{false} = p \).
- This is not a tautology because it can be either true or false depending on \( p \).

- **Option (C):** \( (p \land r) \to (p \lor r) \)
- Here, \( (p \land r) \) simplifies to \( (p \land \text{false}) = \text{false} \).
- Thus, we have \( \text{false} \to (p \lor r) \).
- The implication \( \text{false} \to (p \lor r) \) is always true regardless of the truth value of \( (p \lor r) \). Therefore, this is a tautology.

- **Option (D):** \( p \land r \)
- This simplifies to \( p \land \text{false} = \text{false} \).
- This is not a tautology since it is always false.

3. **Conclusion:**
- The only option that is a tautology is **Option (C)**: \( (p \land r) \to (p \lor r) \).

### Final Answer:
**Option (C)** is the tautology.