For the statements p and q, consider the following compound statements :
(a) `(~q ^^ (p to q)) to ~p`
(b) `((p vv q) ^^ ~p) to q`
Then which of the following statements is correct?

Options

A. (a) and (b) both are not tautologies.
B. (a) and (b) both are tautologies.
C. (a) is a tautology but not (b).
D. (b) is a tautology but not (a).

Answer

To solve the problem, we need to analyze the two compound statements given and determine if they are tautologies. A tautology is a statement that is always true regardless of the truth values of its components.

Let's denote:
- \( p \) and \( q \) are the statements.
- \( \neg p \) is the negation of \( p \).
- \( \neg q \) is the negation of \( q \).
- \( p \to q \) is the implication from \( p \) to \( q \).
- \( p \land q \) is the conjunction (AND) of \( p \) and \( q \).
- \( p \lor q \) is the disjunction (OR) of \( p \) and \( q \).

### Step 1: Create a truth table for \( p \) and \( q \)

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

### Step 2: Evaluate statement (a): \( (\neg q \land (p \to q)) \to \neg p \)

1. **Calculate \( \neg q \land (p \to q) \)**:
- For each combination of \( p \) and \( q \):
- When \( p = T, q = T \): \( \neg q = F \), \( p \to q = T \) → \( F \land T = F \)
- When \( p = T, q = F \): \( \neg q = T \), \( p \to q = F \) → \( T \land F = F \)
- When \( p = F, q = T \): \( \neg q = F \), \( p \to q = T \) → \( F \land T = F \)
- When \( p = F, q = F \): \( \neg q = T \), \( p \to q = T \) → \( T \land T = T \)

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

2. **Now calculate \( (\neg q \land (p \to q)) \to \neg p \)**:
- When \( \neg q \land (p \to q) = F \): \( F \to \neg p \) is true regardless of \( \neg p \).
- When \( \neg q \land (p \to q) = T \): \( T \to \neg p \) depends on \( \neg p \):
- If \( p = F \) → \( \neg p = T \) → \( T \to T = T \)
- If \( p = T \) → \( \neg p = F \) → \( T \to F = F \)

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

Thus, statement (a) is a tautology.

### Step 3: Evaluate statement (b): \( ((p \lor q) \land \neg p) \to q \)

1. **Calculate \( (p \lor q) \land \neg p \)**:
- For each combination of \( p \) and \( q \):
- When \( p = T, q = T \): \( p \lor q = T \), \( \neg p = F \) → \( T \land F = F \)
- When \( p = T, q = F \): \( p \lor q = T \), \( \neg p = F \) → \( T \land F = F \)
- When \( p = F, q = T \): \( p \lor q = T \), \( \neg p = T \) → \( T \land T = T \)
- When \( p = F, q = F \): \( p \lor q = F \), \( \neg p = T \) → \( F \land T = F \)

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

2. **Now calculate \( ((p \lor q) \land \neg p) \to q \)**:
- When \( (p \lor q) \land \neg p = F \): \( F \to q \) is true regardless of \( q \).
- When \( (p \lor q) \land \neg p = T \): \( T \to q \) depends on \( q \):
- If \( q = T \) → \( T \to T = T \)
- If \( q = F \) → \( T \to F = F \)

| \( p \) | \( q \) | \( \neg p \) | \( (p \lor q) \land \neg p \) | \( ((p \lor q) \land \neg p) \to q \) |
|---------|---------|---------------|--------------------------------|---------------------------------------|
| T | T | F | F | T |
| T | F | F | F | T |
| F | T | T | T | T |
| F | F | T | F | T |

Thus, statement (b) is also a tautology.

### Conclusion:
Both statements (a) and (b) are tautologies.