Question 17

First simplify the compound proposition $\neg p \land (p \lor q)$:

Now analyze $S_1: (\neg p \land q) \rightarrow q$. In propositional logic, $(A \land B) \rightarrow B$ is an established tautology (simplification rule). Whenever the antecedent $(\neg p \land q)$ is true, $q$ is necessarily true. Hence, $S_1$ is a tautology.

Now analyze $S_2: q \rightarrow (\neg p \land q)$. Consider the truth assignment where $p = \text{True}$ and $q = \text{True}$. Under this assignment, the antecedent $q$ is true, but the consequent $\neg p \land q = \text{False} \land \text{True} = \text{False}$. Thus $\text{True} \rightarrow \text{False} = \text{False}$, so $S_2$ does not hold universally.

Consequently, $S_1$ is a tautology but $S_2$ is not a tautology, which matches option B.