The statement `prArr(q^^p)` is negation of the statement

Options

A. `prArrq`
B. `p^^q`
C. `~(prArrq)`
D. `~(p^^q)`

Answer

To find the negation of the statement \( S: P \implies Q \land P \), we will follow these steps:

### Step 1: Understand the original statement
The original statement \( S \) is \( P \implies (Q \land P) \). This means "If \( P \) is true, then both \( Q \) and \( P \) are true."

### Step 2: Apply the negation
To find the negation of \( S \), we write:
\[ \neg S = \neg (P \implies (Q \land P)) \]

### Step 3: Rewrite the implication
Using the logical equivalence \( A \implies B \equiv \neg A \lor B \), we can rewrite the implication:
\[ P \implies (Q \land P) \equiv \neg P \lor (Q \land P) \]
Thus, we have:
\[ \neg S = \neg (\neg P \lor (Q \land P)) \]

### Step 4: Apply De Morgan's Law
Using De Morgan's Law, we can negate the expression:
\[ \neg (\neg P \lor (Q \land P)) \equiv P \land \neg (Q \land P) \]

### Step 5: Further simplify using De Morgan's Law
Now we can apply De Morgan's Law again to \( \neg (Q \land P) \):
\[ \neg (Q \land P) \equiv \neg Q \lor \neg P \]
So, we have:
\[ \neg S = P \land (\neg Q \lor \neg P) \]

### Step 6: Distribute \( P \)
Distributing \( P \) gives us:
\[ \neg S = (P \land \neg Q) \lor (P \land \neg P) \]
Since \( P \land \neg P \) is always false, we can simplify this to:
\[ \neg S = P \land \neg Q \]

### Conclusion
Thus, the negation of the statement \( S: P \implies (Q \land P) \) is:
\[ \neg S = P \land \neg Q \]