The Boolean Expression `(p^^~ q)vvqvv(~ p^^q)` is equivalent to :

Options

A. `p^^q`
B. `pvvq`
C. `pvv~p`
D. `~p^^q`

Explanation

To solve the Boolean expression `(p ∧ ¬q) ∨ q ∨ (¬p ∧ q)` and determine its equivalence, we will create a truth table and evaluate the expression step by step.

### Step 1: Define the Variables
We will define the variables involved:
- Let `p` and `q` be the Boolean variables.
- `¬p` is the negation of `p`.
- `¬q` is the negation of `q`.

### Step 2: Create the Truth Table
We will create a truth table that includes all possible values of `p` and `q`, along with their negations and the results of the expressions involved.

| p | q | ¬p | ¬q | p ∧ ¬q | ¬p ∧ q | (p ∧ ¬q) ∨ q | (p ∧ ¬q) ∨ q ∨ (¬p ∧ q) |
|---|---|----|----|--------|---------|---------------|--------------------------|
| T | T | F | F | F | F | T | T |
| T | F | F | T | T | F | T | T |
| F | T | T | F | F | T | T | T |
| F | F | T | T | F | F | F | F |

### Step 3: Evaluate Each Expression
1. **Negations**:
- When `p` is True (T), `¬p` is False (F). When `p` is False (F), `¬p` is True (T).
- When `q` is True (T), `¬q` is False (F). When `q` is False (F), `¬q` is True (T).

2. **Calculate `p ∧ ¬q`**:
- This expression is True only when `p` is True and `q` is False. Thus, it is True in the second row.

3. **Calculate `¬p ∧ q`**:
- This expression is True only when `p` is False and `q` is True. Thus, it is True in the third row.

4. **Calculate `(p ∧ ¬q) ∨ q`**:
- This expression is True if either `p ∧ ¬q` is True or `q` is True. It is True in the first three rows.

5. **Final Expression `(p ∧ ¬q) ∨ q ∨ (¬p ∧ q)`**:
- This expression is True if either `(p ∧ ¬q) ∨ q` is True or `¬p ∧ q` is True. It is True in the first three rows and False only in the last row.

### Step 4: Conclusion
From the truth table, we see that the final column `(p ∧ ¬q) ∨ q ∨ (¬p ∧ q)` evaluates to True for the combinations (T, T), (T, F), (F, T) and False for (F, F). This is equivalent to the expression `p ∨ q`.

### Final Answer
The Boolean expression `(p ∧ ¬q) ∨ q ∨ (¬p ∧ q)` is equivalent to `p ∨ q`.