Statement 1 : The statement `A rarr (A rarr B)`. Is equivalent to `A rarr (A vv B)`
statement 2 : The statement `~[(A ^^ B) rarr (~A vv B)]` is a tautology

Options

A. Statement 1 and statement 2 are both false
B. Statemet 1 and statement 2 are both true
C. statement 1 is true and statement 2 is false
D. statement 1 is false and statement 2 is true

Answer

To solve the problem, we need to analyze both statements and determine their truth values.

### Statement 1:
**Statement:** \( A \implies (A \implies B) \) is equivalent to \( A \implies (A \lor B) \).

**Step 1:** Understand the implication \( A \implies (A \implies B) \).
- The implication \( A \implies (A \implies B) \) can be rewritten using the definition of implication:
\[ A \implies (A \implies B) \equiv A \implies (\neg A \lor B) \]

**Step 2:** Understand the implication \( A \implies (A \lor B) \).
- Similarly, we can rewrite \( A \implies (A \lor B) \):
\[ A \implies (A \lor B) \equiv \neg A \lor (A \lor B) \equiv \neg A \lor A \lor B \]
- By the law of excluded middle, \( \neg A \lor A \) is always true. Thus, \( \neg A \lor (A \lor B) \) is always true.

**Step 3:** Evaluate both expressions.
- We need to check if both expressions yield the same truth values under all combinations of truth values for \( A \) and \( B \).

| A | B | \( A \implies (A \implies B) \) | \( A \implies (A \lor B) \) |
|---|---|-------------------------------|-----------------------------|
| T | T | T | T |
| T | F | F | F |
| F | T | T | T |
| F | F | T | T |

- From the table, we see that both expressions yield the same truth values for all combinations of \( A \) and \( B \).

**Conclusion for Statement 1:**
- Therefore, Statement 1 is **True**.

### Statement 2:
**Statement:** \( \neg[(A \land B) \implies (\neg A \lor B)] \) is a tautology.

**Step 1:** Rewrite the implication in Statement 2.
- The implication \( (A \land B) \implies (\neg A \lor B) \) can be rewritten as:
\[ \neg(A \land B) \lor (\neg A \lor B) \]

**Step 2:** Simplify the expression.
- This can be simplified using De Morgan's laws:
\[ \neg(A \land B) \equiv \neg A \lor \neg B \]
- Therefore, we have:
\[ \neg(A \land B) \lor (\neg A \lor B) \equiv (\neg A \lor \neg B) \lor (\neg A \lor B) \]
- This simplifies to:
\[ \neg A \lor (\neg B \lor B) \equiv \neg A \lor \text{True} \equiv \text{True} \]

**Step 3:** Negate the expression.
- Now, we need to negate the entire expression:
\[ \neg[(A \land B) \implies (\neg A \lor B)] \equiv \neg \text{True} \equiv \text{False} \]

**Conclusion for Statement 2:**
- Since the negation results in **False**, Statement 2 is **False**.

### Final Conclusion:
- Statement 1 is **True**.
- Statement 2 is **False**.