Among the statements :
(S1) `((p vv q)rArr r)hArr(p rarr r)`
(S2) `(p vv q)rArr r)hArr(p rArr r)vv(q rArr r)`

Options

A. only (S2) is a tautology
B. both (S1) and (S2) are tautologies
C. only (S1) is a tautology
D. neither (S1) nor (S2) is a tautology

Answer

To solve the problem, we need to analyze the two statements (S1) and (S2) to determine if they are tautologies or not. A tautology is a statement that is true in every possible interpretation.

### Step 1: Analyze Statement S1
The first statement is:
\[ (S1) \quad ((p \lor q) \implies r) \iff (p \implies r) \]

1. **Break down the components**:
- \(p \lor q\) means "p or q".
- \((p \lor q) \implies r\) means "if p or q is true, then r is true".
- \(p \implies r\) means "if p is true, then r is true".

2. **Construct a truth table** for \(p\), \(q\), and \(r\) to evaluate the truth values of both sides of the equivalence.

| p | q | r | \(p \lor q\) | \((p \lor q) \implies r\) | \(p \implies r\) | \((p \lor q) \implies r \iff (p \implies r)\) |
|---|---|---|-------------|----------------------------|------------------|-----------------------------------------------|
| T | T | T | T | T | T | T |
| T | T | F | T | F | F | T |
| T | F | T | T | T | T | T |
| T | F | F | T | F | F | T |
| F | T | T | T | T | T | T |
| F | T | F | T | F | T | F |
| F | F | T | F | T | T | T |
| F | F | F | F | T | T | T |

3. **Evaluate the results**:
- The last column shows that \((p \lor q) \implies r \iff (p \implies r)\) is not true for all combinations (specifically when \(p\) is false and \(q\) is true, it is false). Thus, S1 is **not a tautology**.

### Step 2: Analyze Statement S2
The second statement is:
\[ (S2) \quad ((p \lor q) \implies r) \iff (p \implies r) \lor (q \implies r) \]

1. **Break down the components**:
- The left side is the same as before: \((p \lor q) \implies r\).
- The right side is \((p \implies r) \lor (q \implies r)\), meaning "either p implies r or q implies r".

2. **Construct a truth table** for this statement as well.

| p | q | r | \(p \lor q\) | \((p \lor q) \implies r\) | \(p \implies r\) | \(q \implies r\) | \((p \implies r) \lor (q \implies r)\) | \((p \lor q) \implies r \iff ((p \implies r) \lor (q \implies r))\) |
|---|---|---|-------------|----------------------------|------------------|------------------|-----------------------------------------|----------------------------------------------------------|
| T | T | T | T | T | T | T | T | T |
| T | T | F | T | F | F | F | F | T |
| T | F | T | T | T | T | T | T | T |
| T | F | F | T | F | F | T | T | F |
| F | T | T | T | T | T | T | T | T |
| F | T | F | T | F | T | F | T | F |
| F | F | T | F | T | T | T | T | T |
| F | F | F | F | T | T | T | T | T |

3. **Evaluate the results**:
- The last column shows that \((p \lor q) \implies r \iff ((p \implies r) \lor (q \implies r))\) is true for all combinations except when \(p\) is true and \(q\) is false with \(r\) false. Thus, S2 is also **not a tautology**.

### Conclusion
Since neither S1 nor S2 is a tautology, the answer to the question is that **neither S1 nor S2 is a tautology**.