`~(prArr q)hArr ~pvv p)` is a

Options

A. Tautology and contradiction
B. a contradiction
C. neither a tautology nor a contradiction
D. cannot come to any conclusion

Answer

To solve the proposition `~(p → q) ↔ ~p ∨ p`, we will use a truth table to evaluate the expression step by step.

### Step 1: Identify the Variables
We have two variables, \( p \) and \( q \). Each can be either true (T) or false (F).

### Step 2: Create the Truth Table
We will create a truth table with all combinations of truth values for \( p \) and \( q \).

| \( p \) | \( q \) | \( p \to q \) | \( \sim (p \to q) \) | \( \sim p \) | \( \sim p \lor p \) | \( \sim (p \to q) \leftrightarrow (\sim p \lor p) \) |
|---------|---------|---------------|----------------------|--------------|---------------------|-----------------------------------------------------|
| T | T | T | F | F | T | F |
| T | F | F | T | F | T | T |
| F | T | T | F | T | T | F |
| F | F | T | F | T | T | F |

### Step 3: Calculate \( p \to q \)
The implication \( p \to q \) is only false when \( p \) is true and \( q \) is false. In all other cases, it is true.

### Step 4: Calculate \( \sim (p \to q) \)
Negate the values of \( p \to q \):
- If \( p \to q \) is T, then \( \sim (p \to q) \) is F.
- If \( p \to q \) is F, then \( \sim (p \to q) \) is T.

### Step 5: Calculate \( \sim p \)
Negate the values of \( p \):
- If \( p \) is T, then \( \sim p \) is F.
- If \( p \) is F, then \( \sim p \) is T.

### Step 6: Calculate \( \sim p \lor p \)
This is a tautology since one of the two will always be true:
- If \( p \) is T, \( \sim p \) is F, so \( \sim p \lor p \) is T.
- If \( p \) is F, \( \sim p \) is T, so \( \sim p \lor p \) is T.

### Step 7: Calculate the Final Expression \( \sim (p \to q) \leftrightarrow (\sim p \lor p) \)
Now we compare \( \sim (p \to q) \) with \( \sim p \lor p \):
- If both are the same (both T or both F), the result is T.
- If they are different, the result is F.

### Final Evaluation
From the truth table, the final column shows that the expression is neither always true (tautology) nor always false (contradiction). It has both true and false values.

### Conclusion
The expression `~(p → q) ↔ ~p ∨ p` is **neither a tautology nor a contradiction**.

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