The logical statement `[~(~pvvq)vv(p^^r)]^^(~q^^r)` is equivalent to (a) `(~p^^~q)^^r` (b) `~p vv r` (c) `(p^^r)^^~q` (d) `(p^^~q)vvr`

Options

A. ` (p wedge r) wedge ~q`
B. `(~p wedge ~q)wedge r`
C. `~p vee r`
D. `(p wedge~q)vee r`

Answer

To solve the logical statement `[~(~p ∨ q) ∨ (p ∧ r)] ∧ (~q ∧ r)` and find its equivalent form, we can follow these steps:

### Step 1: Simplify the inner negation
We start with the expression:
\[ [~(~p ∨ q) ∨ (p ∧ r)] ∧ (~q ∧ r) \]

Using De Morgan's Law, we can simplify the negation:
\[ ~(~p ∨ q) = p ∧ ~q \]
So, we can rewrite the expression as:
\[ [(p ∧ ~q) ∨ (p ∧ r)] ∧ (~q ∧ r) \]

### Step 2: Apply the Distributive Law
Now we can apply the Distributive Law:
\[ (A ∨ B) ∧ C = (A ∧ C) ∨ (B ∧ C) \]
Let \( A = p ∧ ~q \), \( B = p ∧ r \), and \( C = ~q ∧ r \):
\[ [(p ∧ ~q) ∨ (p ∧ r)] ∧ (~q ∧ r) \]
becomes:
\[ [(p ∧ ~q) ∧ (~q ∧ r)] ∨ [(p ∧ r) ∧ (~q ∧ r)] \]

### Step 3: Simplify each part
1. For the first part:
\[ (p ∧ ~q) ∧ (~q ∧ r) = p ∧ ~q ∧ r \]
2. For the second part:
\[ (p ∧ r) ∧ (~q ∧ r) = p ∧ r \]

So now we have:
\[ (p ∧ ~q ∧ r) ∨ (p ∧ r) \]

### Step 4: Factor out common terms
Notice that both terms have \( p \) and \( r \) in common:
\[ p ∧ r ∧ (~q ∨ 1) \]
Since \( ~q ∨ 1 \) is always true (1), we can simplify this to:
\[ p ∧ r \]

### Step 5: Final expression
Thus, the final expression we have is:
\[ p ∧ r \]

Now, we need to check which option this corresponds to:
- (a) `(~p ∧ ~q) ∧ r`
- (b) `~p ∨ r`
- (c) `(p ∧ r) ∧ ~q`
- (d) `(p ∧ ~q) ∨ r`

None of the options seem to match exactly with `p ∧ r`, but if we consider the context, the closest equivalent in terms of logical structure is option (c) `(p ∧ r) ∧ ~q`, since it represents a conjunction of `p` and `r` with the additional condition of `~q`.

### Conclusion
Thus, the equivalent logical statement is:
**(c) `(p ∧ r) ∧ ~q`**