`~ (p implies q) hArr ~ p vv ~ q` is-

Options

A. a tautology
B. a contraction
C. neither a tautology nor a contradiction
D. cannot come to any conciusion

Answer

To determine whether the statement `~ (p implies q) ↔ ~p ∨ ~q` is a tautology, contradiction, or neither, we will create a truth table and analyze the results step by step.

### Step 1: Define the Implication
The implication \( p \implies q \) can be rewritten in terms of logical operations:
\[ p \implies q \equiv \neg p \lor q \]
This means that \( p \implies q \) is false only when \( p \) is true and \( q \) is false.

### Step 2: Create the Truth Table
We will create a truth table for \( p \) and \( q \), and then calculate \( p \implies q \), \( \neg (p \implies q) \), \( \neg p \), \( \neg q \), and \( \neg p \lor \neg q \).

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

### Step 3: Calculate Negation of Implication
From the truth table, we see:
- When \( p \) is true and \( q \) is true, \( p \implies q \) is true, so \( \neg (p \implies q) \) is false.
- When \( p \) is true and \( q \) is false, \( p \implies q \) is false, so \( \neg (p \implies q) \) is true.
- When \( p \) is false and \( q \) is true, \( p \implies q \) is true, so \( \neg (p \implies q) \) is false.
- When \( p \) is false and \( q \) is false, \( p \implies q \) is true, so \( \neg (p \implies q) \) is false.

### Step 4: Calculate Negation of p and q
From the truth table:
- \( \neg p \) is true when \( p \) is false and false when \( p \) is true.
- \( \neg q \) is true when \( q \) is false and false when \( q \) is true.

### Step 5: Calculate the OR Condition
Now we calculate \( \neg p \lor \neg q \):
- This expression is true if either \( \neg p \) or \( \neg q \) is true.

### Step 6: Evaluate the Biconditional
Now we need to evaluate the biconditional \( \neg (p \implies q) \leftrightarrow (\neg p \lor \neg q) \):
- The biconditional is true if both sides are either true or both are false.

| \( \neg (p \implies q) \) | \( \neg p \lor \neg q \) | Biconditional Result |
|----------------------------|---------------------------|-----------------------|
| F | F | T |
| T | T | T |
| F | T | F |
| F | T | F |

### Step 7: Conclusion
The final column shows that the biconditional is true in some cases and false in others. Therefore, the statement `~ (p implies q) ↔ ~p ∨ ~q` is neither a tautology (which would require it to be true in all cases) nor a contradiction (which would require it to be false in all cases).

Thus, the answer is that the statement is **neither a tautology nor a contradiction**.