Given the following two statements
`S_(1) : (p ^^ : p) rarr (p ^^ q)` is a tautology.
`S_(2) : (p vv : p) rarr (p vv q)` is a fallacy

Options

A. `S_(1)` is true
B. `S_(2)` is true
C. Both `S_(1)` and `S_(2)` are true
D. None of these

Explanation

To solve the problem, we need to analyze the two statements given:

1. **Statement S1**: \((p \land \neg p) \rightarrow (p \land q)\)
2. **Statement S2**: \((p \lor \neg p) \rightarrow (p \lor q)\)

We will determine whether each statement is a tautology or a fallacy by constructing truth tables for both.

### Step 1: Analyze Statement S1

**Statement S1**: \((p \land \neg p) \rightarrow (p \land q)\)

- **Truth Table Construction**:

| p | ¬p | p ∧ ¬p | q | p ∧ q | (p ∧ ¬p) → (p ∧ q) |
|-------|-------|--------|-------|-------|---------------------|
| T | F | F | T | T | T |
| T | F | F | F | F | T |
| F | T | F | T | F | T |
| F | T | F | F | F | T |

- **Explanation**:
- The column for \(p \land \neg p\) is always false (F) because \(p\) cannot be true and false at the same time.
- Since \(p \land \neg p\) is false, the implication \((p \land \neg p) \rightarrow (p \land q)\) is always true (T) regardless of the truth values of \(p\) and \(q\).

- **Conclusion**: Statement S1 is a tautology.

### Step 2: Analyze Statement S2

**Statement S2**: \((p \lor \neg p) \rightarrow (p \lor q)\)

- **Truth Table Construction**:

| p | ¬p | p ∨ ¬p | q | p ∨ q | (p ∨ ¬p) → (p ∨ q) |
|-------|-------|--------|-------|-------|---------------------|
| T | F | T | T | T | T |
| T | F | T | F | T | T |
| F | T | T | T | T | T |
| F | T | T | F | F | F |

- **Explanation**:
- The column for \(p \lor \neg p\) is always true (T) because either \(p\) is true or \(\neg p\) is true.
- The implication \((p \lor \neg p) \rightarrow (p \lor q)\) is false (F) when \(p\) is false and \(q\) is also false.

- **Conclusion**: Statement S2 is a fallacy since it is not true in all cases.

### Final Conclusion

- **S1** is a tautology.
- **S2** is a fallacy.