1. **State the problem:** We want to show that the statement $\sim r \to s$ logically follows from the premises $p \to r$, $\sim p \to q$, and $q \to s$. 2. **Analyze the premises:** - Premise 1: $p \to r$ means if $p$ is true, then $r$ is true. - Premise 2: $\sim p \to q$ means if $p$ is false, then $q$ is true. - Premise 3: $q \to s$ means if $q$ is true, then $s$ is true. 3. **Goal:** Show $\sim r \to s$, i.e., if $r$ is false, then $s$ is true. 4. **Proof by cases:** Consider the truth value of $p$. - Case 1: Suppose $p$ is true. - From $p \to r$, since $p$ is true, $r$ must be true. - But this contradicts the assumption $\sim r$ (i.e., $r$ is false). - So this case cannot happen if $\sim r$ is true. - Case 2: Suppose $p$ is false. - From $\sim p \to q$, since $p$ is false, $q$ is true. - From $q \to s$, since $q$ is true, $s$ is true. 5. **Conclusion:** If $\sim r$ is true, then $p$ cannot be true (from case 1), so $p$ is false. From case 2, if $p$ is false, then $s$ is true. Therefore, $\sim r \to s$ logically follows from the premises. **Final answer:** $\boxed{\sim r \to s}$