If p and q are logical statements, then `(~p)rarr(prarrq)` is equivalent to

Options

A. `p^^q`
B. `prarr(pvvq)`
C. `pvvq`
D. None of these

Answer

To solve the logical statement `(~p) → (p → q)` and find its equivalent form, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Implication**:
The statement `(~p) → (p → q)` can be rewritten using the definition of implication. Recall that `A → B` is equivalent to `¬A ∨ B`. Therefore, we can rewrite the implication:
\[ (~p) → (p → q) \equiv ¬(~p) ∨ (p → q) \]

2. **Negate the Negation**:
The negation of `~p` is `p`. So we can substitute this into our expression:
\[ p ∨ (p → q) \]

3. **Rewrite the Inner Implication**:
Now, we need to rewrite `p → q` using the same implication rule:
\[ p → q \equiv ¬p ∨ q \]
Substituting this back into our expression gives:
\[ p ∨ (¬p ∨ q) \]

4. **Apply Associative Law**:
The logical OR operation is associative, which allows us to rearrange the terms:
\[ p ∨ ¬p ∨ q \]

5. **Use the Law of Excluded Middle**:
The expression `p ∨ ¬p` is a tautology, meaning it is always true. Therefore, we can simplify:
\[ (p ∨ ¬p) ∨ q \equiv True ∨ q \]

6. **Final Simplification**:
Since `True ∨ q` is always true regardless of the truth value of `q`, we conclude:
\[ True \]

### Conclusion:
Thus, the expression `(~p) → (p → q)` is equivalent to `True`.