Negation of the statement `(p ^^ r) -> (r vv q) ` is-

Options

A. ` ( p ^^ r) ^^ ( ~ r ^^ ~ q)`
B. ` ~ ( p ^^ r) to ~ ( r vv q)`
C. ` ~ ( p vv r) to ~ ( r ^^ q)`
D. ` ( p ^^q) vv ( r vv q)`

Explanation

To find the negation of the statement \( (p \land r) \to (r \lor q) \), we will follow these steps:

### Step 1: Understand the original statement
The original statement is an implication of the form \( A \to B \), where:
- \( A \) is \( (p \land r) \)
- \( B \) is \( (r \lor q) \)

### Step 2: Use the negation of an implication
The negation of an implication \( A \to B \) is given by the formula:
\[ \neg(A \to B) \equiv A \land \neg B \]
This means we need to find \( A \) and \( \neg B \).

### Step 3: Find \( A \) and \( \neg B \)
We already have:
- \( A = (p \land r) \)

Now, we need to find \( \neg B \):
- \( B = (r \lor q) \)
- Therefore, \( \neg B = \neg(r \lor q) \)

Using De Morgan's laws, we can express \( \neg B \) as:
\[ \neg(r \lor q) \equiv \neg r \land \neg q \]

### Step 4: Combine \( A \) and \( \neg B \)
Now we can substitute back into the negation of the implication:
\[ \neg((p \land r) \to (r \lor q)) \equiv (p \land r) \land (\neg r \land \neg q) \]

### Step 5: Simplify the expression
We can rewrite the expression:
\[ (p \land r) \land (\neg r \land \neg q) \equiv (p \land r \land \neg r) \land \neg q \]

Since \( r \land \neg r \) is a contradiction (always false), we can simplify this to:
\[ \text{False} \land \neg q \equiv \text{False} \]

### Final Answer
Thus, the negation of the statement \( (p \land r) \to (r \lor q) \) is:
\[ \text{False} \]