Negation of the proposition `(p vv q) ^^(~q^^r)` is

Options

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

Answer

To find the negation of the proposition \((p \lor q) \land (\neg q \land r)\), we will follow the rules of logical negation step by step.

### Step 1: Identify the proposition
The given proposition is:
\[ (p \lor q) \land (\neg q \land r) \]

### Step 2: Apply negation
To negate the entire proposition, we apply the negation operator:
\[ \neg((p \lor q) \land (\neg q \land r)) \]

### Step 3: Use De Morgan's Laws
According to De Morgan's Laws, the negation of a conjunction is the disjunction of the negations:
\[ \neg(A \land B) = \neg A \lor \neg B \]
Here, \(A\) is \((p \lor q)\) and \(B\) is \((\neg q \land r)\). Thus, we can rewrite the negation as:
\[ \neg(p \lor q) \lor \neg(\neg q \land r) \]

### Step 4: Negate each part
Now we will negate each part separately.

1. Negate \(p \lor q\):
\[ \neg(p \lor q) = \neg p \land \neg q \quad \text{(by De Morgan's Laws)} \]

2. Negate \(\neg q \land r\):
\[ \neg(\neg q \land r) = \neg(\neg q) \lor \neg r = q \lor \neg r \quad \text{(by De Morgan's Laws)} \]

### Step 5: Combine the results
Now we substitute back into our expression:
\[ \neg(p \lor q) \lor \neg(\neg q \land r) = (\neg p \land \neg q) \lor (q \lor \neg r) \]

### Step 6: Simplify if necessary
The expression can be left as is, but for clarity, we can write it as:
\[ (\neg p \land \neg q) \lor (q \lor \neg r) \]

### Final Result
Thus, the negation of the proposition \((p \lor q) \land (\neg q \land r)\) is:
\[ (\neg p \land \neg q) \lor (q \lor \neg r) \]