The Boolean expression `~(p rArr (~q))` is equivalent to:

Options

A. `(~p)impliesq`
B. `pvvq`
C. `p^^q`
D. `qimplies~q`

Explanation

To simplify the Boolean expression `~(p → (~q))`, we can follow these steps:

### Step 1: Rewrite the implication
The implication `p → (~q)` can be rewritten using the equivalence `p → q ≡ ~p ∨ q`. Therefore, we can express `p → (~q)` as:
\[ p → (~q) \equiv ~p ∨ (~q) \]

### Step 2: Apply negation
Now, we need to apply the negation to the expression we obtained in Step 1:
\[ ~(p → (~q)) \equiv ~(~p ∨ (~q)) \]

### Step 3: Use De Morgan's Law
According to De Morgan's Law, the negation of a disjunction is equivalent to the conjunction of the negations:
\[ ~(A ∨ B) \equiv ~A ∧ ~B \]
Applying this to our expression:
\[ ~(~p ∨ (~q)) \equiv ~~p ∧ ~~(~q) \]

### Step 4: Simplify double negation
The double negation can be simplified:
\[ ~~p ∧ ~~(~q) \equiv p ∧ q \]

Thus, the simplified expression is:
\[ p ∧ q \]

### Final Answer
The Boolean expression `~(p → (~q))` is equivalent to:
\[ p ∧ q \]