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 both statements \( S_1 \) and \( S_2 \) to determine their validity.

### Step 1: Analyze Statement \( S_1 \)
Statement \( S_1 \): \( (p \land \neg p) \rightarrow (p \land q) \)

1. **Understanding \( p \land \neg p \)**:
- The expression \( p \land \neg p \) is a contradiction. It can never be true because \( p \) cannot be both true and false at the same time.
- Therefore, \( p \land \neg p \) is always false.

2. **Implication Truth Table**:
- The implication \( A \rightarrow B \) is only false when \( A \) is true and \( B \) is false. In all other cases, it is true.
- Since \( p \land \neg p \) is always false, the implication \( (p \land \neg p) \rightarrow (p \land q) \) will always be true regardless of the truth values of \( p \) and \( q \).

Thus, \( S_1 \) is a tautology.

### Step 2: Analyze Statement \( S_2 \)
Statement \( S_2 \): \( (p \lor \neg p) \rightarrow (p \lor q) \)

1. **Understanding \( p \lor \neg p \)**:
- The expression \( p \lor \neg p \) is a tautology. It is always true because either \( p \) is true or \( \neg p \) is true.

2. **Implication Truth Table**:
- Again, using the implication truth table, since \( p \lor \neg p \) is always true, the implication \( (p \lor \neg p) \rightarrow (p \lor q) \) will depend on the truth value of \( p \lor q \).
- If \( p \lor q \) is true (which can happen in several cases), then the implication is true. If \( p \lor q \) is false, then the implication is false.

3. **Evaluating \( p \lor q \)**:
- The expression \( p \lor q \) can be false only when both \( p \) and \( q \) are false.
- Therefore, \( S_2 \) is not a fallacy because it can be true in cases where \( p \) or \( q \) is true.

Thus, \( S_2 \) is not a fallacy.

### Conclusion
- \( S_1 \) is a tautology.
- \( S_2 \) is not a fallacy.

The correct option is that \( S_1 \) is true and \( S_2 \) is false.