Syllogism — Choose which conclusions must be true: Statements: • All books are trees. • All trees are lions. Conclusions: I. All books are lions. II. All lions are books. III. All trees are books. IV. Some lions are books.

Introduction / Context:
This is a textbook “chain” problem with two universal affirmatives. You must recognize transitivity of inclusion and then consider whether an existential claim (“some …”) is warranted.

Given Data / Assumptions:

• S1: Books ⊆ Trees.
• S2: Trees ⊆ Lions.
• Conclusions: I “Books ⊆ Lions”; II “Lions ⊆ Books”; III “Trees ⊆ Books”; IV “Some Lions are Books.”

Concept / Approach:
From S1 and S2, Books ⊆ Lions by transitivity. The converses II and III are not implied. For IV, most exam conventions assume non-emptiness of the named classes; if any book exists, that book is a lion, yielding “Some lions are books.”

Step-by-Step Solution:
1) S1 + S2 ⇒ Books ⊆ Lions. So I follows.2) II (“All lions are books”) would need Lions ⊆ Books, not provided.3) III (“All trees are books”) would need Trees ⊆ Books, also not provided.4) If Books is non-empty (standard assumption), then ∃x Book(x). From S1 and S2, Book(x) ⇒ Lion(x), proving IV.

Verification / Alternative check:
Draw Books inside Trees, Trees inside Lions. The diagram makes I and IV immediate and shows II and III need not hold.

Why Other Options Are Wrong:
Any choice including II or III asserts a converse not supported by the premises; “none” ignores I (and often IV).

Common Pitfalls:
Assuming equivalence from inclusion and forgetting to verify existence for “some” conclusions.

Final Answer:
Only Conclusions I and IV follow.