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

Options

A. `p^^q`
B. `prarr(pvvq)`
C. `pvvq`
D. `(pvvq)hArr (p^^q)`

Answer

To determine the logical equivalence of the statement \((\neg p) \rightarrow (p \rightarrow q)\), we will follow these steps:

### Step 1: Understand the Components
The expression consists of:
- \(\neg p\): the negation of \(p\)
- \(p \rightarrow q\): the implication from \(p\) to \(q\)

### Step 2: Construct the Truth Table
We will create a truth table for the expression \((\neg p) \rightarrow (p \rightarrow q)\).

| \(p\) | \(q\) | \(\neg p\) | \(p \rightarrow q\) | \((\neg p) \rightarrow (p \rightarrow q)\) |
|-------|-------|------------|----------------------|---------------------------------------------|
| T | T | F | T | T |
| T | F | F | F | T |
| F | T | T | T | T |
| F | F | T | T | T |

### Step 3: Analyze the Truth Table
From the truth table, we see that the last column, which represents \((\neg p) \rightarrow (p \rightarrow q)\), is true in all cases. This means that the expression is a tautology.

### Step 4: Compare with Options
Now we need to check the options provided to see which one matches this truth table.

1. **Option A**: \(p \land q\) - This is not a tautology as it is false when either \(p\) or \(q\) is false.
2. **Option B**: \(p \rightarrow (p \lor q)\) - This is a tautology as it is true for all combinations of \(p\) and \(q\).
3. **Option C**: \(p \land q\) - This is the same as option A and is not a tautology.
4. **Option D**: \(p \land q\) - This is also the same as options A and C and is not a tautology.

### Conclusion
The only option that matches the truth table of \((\neg p) \rightarrow (p \rightarrow q)\) is **Option B**.

### Final Answer
Thus, \((\neg p) \rightarrow (p \rightarrow q)\) is equivalent to **Option B**.

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