The negation of statement `(~p vvq) ^^ (~p ^^~q)` is -

Options

A. `(p vv -q) ^^ (p vvq)`
B. `(p ^^ ~q) vv (p vvq)`
C. `(~p vv q)vv (~ p ^^ q)`
D. `(p ^^ ~ q) ^^ (p vv q)`

Answer

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

### Step 1: Identify the original statement
The original statement is:
\[ (\neg p \lor q) \land (\neg p \land \neg q) \]

### Step 2: Apply De Morgan's Laws
To negate the entire statement, we apply De Morgan's Laws, which state that:
\[ \neg(A \land B) = \neg A \lor \neg B \]
and
\[ \neg(A \lor B) = \neg A \land \neg B \]

In our case, we have:
\[ \neg((\neg p \lor q) \land (\neg p \land \neg q)) = \neg(\neg p \lor q) \lor \neg(\neg p \land \neg q) \]

### Step 3: Negate each part
Now we will negate each part separately.

1. For \(\neg(\neg p \lor q)\):
- Using De Morgan's Law:
\[ \neg(\neg p \lor q) = \neg(\neg p) \land \neg q = p \land \neg q \]

2. For \(\neg(\neg p \land \neg q)\):
- Again using De Morgan's Law:
\[ \neg(\neg p \land \neg q) = \neg(\neg p) \lor \neg(\neg q) = p \lor q \]

### Step 4: Combine the results
Now we combine the results from the negations:
\[ \neg((\neg p \lor q) \land (\neg p \land \neg q)) = (p \land \neg q) \lor (p \lor q) \]

### Step 5: Simplify if possible
The expression \((p \land \neg q) \lor (p \lor q)\) can be simplified further:
\[ = p \lor (p \land \neg q) \lor q \]
Using the absorption law, \(p \lor (p \land \neg q) = p\):
\[ = p \lor q \]

### Final Answer
Thus, the negation of the statement \((\neg p \lor q) \land (\neg p \land \neg q)\) is:
\[ p \lor q \]