Statements: • Many books are rocks. • All rocks are clips. Conclusions: I. Some books are clips. II. No rock is a book. Choose the option that must follow.

Introduction / Context:
Here a quantified existential (“many” which implies “some”) combines with a universal inclusion. We must transfer membership through the inclusion and also test a negation claim against the premises.

Given Data / Assumptions:

• “Many Books are Rocks” ⇒ at least one book is a rock (∃x ∈ Books ∩ Rocks).
• All Rocks ⊆ Clips.

Concept / Approach:
If some book is a rock and every rock is a clip, then that specific book is also a clip. Conversely, “No rock is a book” contradicts the existential information provided by “many books are rocks.”

Step-by-Step Solution:
1) Pick witness b1 ∈ Books ∩ Rocks.2) From Rocks ⊆ Clips, b1 ∈ Clips. Therefore, Some Books are Clips ⇒ Conclusion I holds.3) Conclusion II (“No rock is a book”) is incompatible with b1 ∈ Books ∩ Rocks and is therefore false.

Verification / Alternative check:
Model: Rocks = {r1, r2}, Books = {b1} with b1 = r1, Clips = {r1, r2, c1}. All premises hold. Conclusion I is satisfied by b1, while II is negated by the same witness.

Why Other Options Are Wrong:
“Both” is impossible (II contradicts the premises). “Only II” contradicts the existential. “Neither” ignores the logical consequence.

Common Pitfalls:
Overlooking that “many” at least means “some,” which is enough to carry through universal inclusions.

Final Answer:
Only Conclusion I follows.