Syllogism – Evaluate conclusions from partial overlaps Statements: • Some hills are rivers. • Some rivers are deserts. • All deserts are roads. Conclusions to test: I. Some roads are rivers. II. Some roads are hills. III. Some deserts are hills.

Introduction / Context:
Syllogism questions often combine “some” (existential overlap) with an “all” (subset) statement. The challenge is to propagate definite implications while resisting the urge to assume extra intersections that are not forced by the data.

Given Data / Assumptions:

• Some hills are rivers (H ∩ R ≠ ∅).
• Some rivers are deserts (R ∩ D ≠ ∅).
• All deserts are roads (D ⊆ Rd).
• Conclusions about “some” require a guaranteed witness element from the premises.

Concept / Approach:
Use set inclusion and overlaps: from R ∩ D and D ⊆ Rd, we can chain membership to show at least one element is both a river and a road. But to connect hills with deserts or roads, we would need a shared element proven to be in both overlaps, which is not provided.

Step-by-Step Solution:
Conclusion I: From “some rivers are deserts” and “all deserts are roads,” the specific elements that are both river and desert are also roads. Therefore, “some roads are rivers” is guaranteed. I follows.Conclusion II: To claim “some roads are hills,” we would need an element shown to be both a hill and a road. We have H ∩ R ≠ ∅ and R ∩ D ≠ ∅, but the hill–river elements could be different from the river–desert elements. No necessary overlap among hills and deserts is given. II does not follow.Conclusion III: “Some deserts are hills” would require D ∩ H ≠ ∅. No statement links deserts directly with hills; it is not guaranteed. III does not follow.

Verification / Alternative check:
Construct a model where one subset of rivers overlaps hills and a different subset overlaps deserts, with these subsets disjoint. All premises hold while II and III fail, confirming only I is compelled.

Why Other Options Are Wrong:

• “All follow” and combinations including II or III assume overlaps not stated.
• “None follows” ignores the clear chain R ∩ D and D ⊆ Rd that yields I.

Common Pitfalls:
Illicitly merging different “some” groups; assuming that if two groups each overlap a third, they overlap each other.

Final Answer:
Only I follows