Among the statements:
S1: $$p \vee q \Rightarrow r \Leftrightarrow p \Rightarrow r$$
S2: $$p \vee q \Rightarrow r \Leftrightarrow p \Rightarrow r \vee q \Rightarrow r$$

Step 1: Understand the Logical Equivalence Rules

The standard logical identity for an implication with a disjunction (OR) in its antecedent is given by:

$$(p \vee q) \Rightarrow r \equiv (p \Rightarrow r) \wedge (q \Rightarrow r)$$

Step 2: Analyze Statement S1

$$S1: ((p \vee q) \Rightarrow r) \Leftrightarrow (p \Rightarrow r)$$

Let us construct a counterexample using truth values to see if this is a tautology:

- Let $$p = \text{False}$$

- Let $$q = \text{True}$$

- Let $$r = \text{False}$$

Evaluate LHS: $$(p \vee q) \Rightarrow r$$

- $$p \vee q = \text{False} \vee \text{True} = \text{True}$$

- $$\text{True} \Rightarrow \text{False} = \text{False}$$

Evaluate RHS: $$p \Rightarrow r$$

- $$\text{False} \Rightarrow \text{False} = \text{True}$$

Since $$\text{LHS} = \text{False}$$ and $$\text{RHS} = \text{True}$$, their bi-conditional equivalence ($$\Leftrightarrow$$) evaluates to $$\text{False}$$.

Therefore, $$S1$$ is not a tautology.

Step 3: Analyze Statement S2

$$S2: ((p \vee q) \Rightarrow r) \Leftrightarrow ((p \Rightarrow r) \vee (q \Rightarrow r))$$

Using our known logical identity from Step 1, the true distribution rule uses an AND ($$\wedge$$) operator, not an OR ($$\vee$$) operator. Let us test with another counterexample:

- Let $$p = \text{True}$$

- Let $$q = \text{False}$$

- Let $$r = \text{False}$$

Evaluate LHS: $$(p \vee q) \Rightarrow r$$

- $$p \vee q = \text{True} \vee \text{False} = \text{True}$$

- $$\text{True} \Rightarrow \text{False} = \text{False}$$

Evaluate RHS: $$(p \Rightarrow r) \vee (q \Rightarrow r)$$

- $$p \Rightarrow r = \text{True} \Rightarrow \text{False} = \text{False}$$

- $$q \Rightarrow r = \text{False} \Rightarrow \text{False} = \text{True}$$

- $$\text{False} \vee \text{True} = \text{True}$$

Since $$\text{LHS} = \text{False}$$ and $$\text{RHS} = \text{True}$$, their bi-conditional equivalence ($$\Leftrightarrow$$) evaluates to $$\text{False}$$.

Therefore, $$S2$$ is also not a tautology.

Conclusion:

Neither (S1) nor (S2) is a tautology.