The compound statement `(P vv Q) ^^ (~P) rArr Q` is equivalent to

Options

A. `P vv Q`
B. `P ^^ ~Q`
C. `~(P rArr Q)`
D. `~(P rArr Q) hArr P ^^ ~Q`

Answer

To determine the equivalence of the compound statement `(P ∨ Q) ∧ (¬P) → Q`, we will analyze it step by step using truth tables.

### Step 1: Identify the components
We have:
- P
- Q
- ¬P (negation of P)
- P ∨ Q (disjunction of P and Q)
- (P ∨ Q) ∧ (¬P) (conjunction of P ∨ Q and ¬P)
- (P ∨ Q) ∧ (¬P) → Q (implication of the previous conjunction to Q)

### Step 2: Create a truth table
We will create a truth table for all combinations of truth values for P and Q.

| P | Q | ¬P | P ∨ Q | (P ∨ Q) ∧ (¬P) | (P ∨ Q) ∧ (¬P) → Q |
|-------|-------|-------|-------|-----------------|---------------------|
| T | T | F | T | F | T |
| T | F | F | T | F | T |
| F | T | T | T | T | T |
| F | F | T | F | F | T |

### Step 3: Fill in the truth table
1. **Negation of P (¬P)**:
- If P is True (T), ¬P is False (F).
- If P is False (F), ¬P is True (T).

2. **Disjunction (P ∨ Q)**:
- True if at least one of P or Q is True.
- Fill in the values based on the truth values of P and Q.

3. **Conjunction ((P ∨ Q) ∧ (¬P))**:
- True only if both (P ∨ Q) and ¬P are True.

4. **Implication ((P ∨ Q) ∧ (¬P) → Q)**:
- This is True unless the first part is True and the second part (Q) is False.

### Step 4: Analyze the results
From the truth table, we see that the column for `(P ∨ Q) ∧ (¬P) → Q` has the following values:
- T
- T
- T
- T

This means that the statement `(P ∨ Q) ∧ (¬P) → Q` is always True regardless of the truth values of P and Q.

### Step 5: Conclusion
Since the statement is always True, it is equivalent to a tautology.