Let P and Q be two statements, then `~(~P^^Q)^^(PvvQ)` is logically equivalent to

Options

A. `Q`
B. `P`
C. `PvvQ`
D. `P^^Q`

Explanation

To solve the expression `~(~P ∧ Q) ∧ (P ∨ Q)`, we will follow a step-by-step approach using logical equivalences.

### Step 1: Rewrite the Expression
We start with the expression:
\[ ~(~P \land Q) \land (P \lor Q) \]

### Step 2: Apply Double Negation
Using the double negation law, we can simplify `~(~P)` to `P`. Thus, we have:
\[ (P \lor Q) \land (P \lor Q) \]

### Step 3: Apply De Morgan's Law
Now, we can apply De Morgan's Law to the negation of the conjunction:
\[ ~(~P \land Q) = P \lor ~Q \]
So, we rewrite our expression as:
\[ (P \lor ~Q) \land (P \lor Q) \]

### Step 4: Distribute the Terms
Now, we can distribute the terms:
\[ (P \lor ~Q) \land (P \lor Q) = P \lor (Q \land ~Q) \]

### Step 5: Simplify Using the Law of Non-Contradiction
Since \( Q \land ~Q \) is always false (denoted as \( \emptyset \)), we simplify this to:
\[ P \lor \emptyset = P \]

### Final Result
Thus, the original expression `~(~P ∧ Q) ∧ (P ∨ Q)` is logically equivalent to:
\[ P \]