Among the two statements
$$(S_1) : (p \Rightarrow q) \wedge (p \wedge (\sim q))$$ is a contradiction and
$$(S_2) : (p \wedge q) \vee ((\sim p) \wedge q) \vee (p \wedge (\sim q)) \vee ((\sim p) \wedge (\sim q))$$ is a tautology

We need to determine which of the statements $$S_1$$ and $$S_2$$ are true.

To analyze $$S_1$$, observe that $$(p \Rightarrow q) \wedge (p \wedge (\sim q))$$ is claimed to be a contradiction.

Recall that $$p \Rightarrow q \equiv (\sim p) \vee q$$. Thus:

$$S_1 = ((\sim p) \vee q) \wedge (p \wedge (\sim q))$$

$$= ((\sim p) \vee q) \wedge p \wedge (\sim q)$$

Distributing yields:

$$= ((\sim p) \wedge p \wedge (\sim q)) \vee (q \wedge p \wedge (\sim q))$$

Since $$(\sim p) \wedge p = F$$ and $$q \wedge (\sim q) = F$$, it follows that:

$$S_1 = F \vee F = F$$

Thus, $$S_1$$ is always false, so it is a contradiction and the claim in $$S_1$$ is true.

Next, consider $$S_2$$, whose expression $$(p \wedge q) \vee ((\sim p) \wedge q) \vee (p \wedge (\sim q)) \vee ((\sim p) \wedge (\sim q))$$ is to be shown a tautology.

We can factor this expression as:

$$= q \wedge (p \vee (\sim p)) \vee (\sim q) \wedge (p \vee (\sim p))$$

Since $$p \vee (\sim p) = T$$, we obtain:

$$= (q \wedge T) \vee ((\sim q) \wedge T) = q \vee (\sim q) = T$$

Hence, $$S_2$$ is always true, i.e., it is a tautology, and the claim in $$S_2$$ is true.

Since both $$S_1$$ and $$S_2$$ are true, the correct answer is Option D: both are true.