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

Options

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

Answer

To solve the given compound statement \( \neg(P \land Q) \lor (\neg P \land Q) \implies (\neg P \land \neg Q) \), we will simplify it step by step.

### Step 1: Rewrite the Implication
The implication \( A \implies B \) can be rewritten as \( \neg A \lor B \). Therefore, we rewrite the statement as:
\[ \neg(\neg(P \land Q) \lor (\neg P \land Q)) \lor (\neg P \land \neg Q) \]

### Step 2: Apply De Morgan's Law
Now, we apply De Morgan's Law to the negation of the left side:
\[ \neg(\neg(P \land Q)) \land \neg(\neg P \land Q) \lor (\neg P \land \neg Q) \]
This simplifies to:
\[ (P \land Q) \land (\neg(\neg P) \lor \neg Q) \lor (\neg P \land \neg Q) \]
Which simplifies further to:
\[ (P \land Q) \land (P \lor \neg Q) \lor (\neg P \land \neg Q) \]

### Step 3: Distribute the Terms
Now we distribute \( (P \land Q) \) over \( (P \lor \neg Q) \):
\[ (P \land Q \land P) \lor (P \land Q \land \neg Q) \lor (\neg P \land \neg Q) \]
Since \( P \land Q \land P \) is just \( P \land Q \), and \( P \land Q \land \neg Q \) is always false (as \( Q \) and \( \neg Q \) cannot be true at the same time), we have:
\[ (P \land Q) \lor (\neg P \land \neg Q) \]

### Step 4: Final Expression
The expression \( (P \land Q) \lor (\neg P \land \neg Q) \) is a well-known logical equivalence. It represents the logical equivalence of \( P \) and \( Q \) being both true or both false. This can be expressed as:
\[ P \equiv Q \]

### Conclusion
Thus, the compound statement \( \neg(P \land Q) \lor (\neg P \land Q) \implies (\neg P \land \neg Q) \) is equivalent to:
\[ P \equiv Q \]