If p and q are two statements, then `(p rArr q) iff (~q rArr ~ p)` is

Options

A. contradiction
B. tautology
C. Neither (a) nor (b)
D. None of these

Explanation

To determine whether the statement \((p \Rightarrow q) \iff (\neg q \Rightarrow \neg p)\) is a tautology, contradiction, or neither, we will create a truth table. Let's break down the steps:

### Step 1: Create a Truth Table for \(p\) and \(q\)

We need to list all possible truth values for the statements \(p\) and \(q\). There are four combinations:

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

### Step 2: Calculate \(p \Rightarrow q\)

The implication \(p \Rightarrow q\) is true in all cases except when \(p\) is true and \(q\) is false.

| \(p\) | \(q\) | \(p \Rightarrow q\) |
|-------|-------|---------------------|
| T | T | T |
| T | F | F |
| F | T | T |
| F | F | T |

### Step 3: Calculate \(\neg q\) and \(\neg p\)

Next, we find the negations of \(q\) and \(p\):

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

### Step 4: Calculate \(\neg q \Rightarrow \neg p\)

Now we calculate the implication \(\neg q \Rightarrow \neg p\):

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

### Step 5: Calculate the Equivalence \((p \Rightarrow q) \iff (\neg q \Rightarrow \neg p)\)

Finally, we check the equivalence of \(p \Rightarrow q\) and \(\neg q \Rightarrow \neg p\):

| \(p\) | \(q\) | \(p \Rightarrow q\) | \(\neg q \Rightarrow \neg p\) | \((p \Rightarrow q) \iff (\neg q \Rightarrow \neg p)\) |
|-------|-------|---------------------|---------------------------------|-------------------------------------------------------|
| T | T | T | T | T |
| T | F | F | F | T |
| F | T | T | T | T |
| F | F | T | T | T |

### Conclusion

The final column shows that \((p \Rightarrow q) \iff (\neg q \Rightarrow \neg p)\) is true for all combinations of \(p\) and \(q\). Therefore, the statement is a **tautology**.