Syllogism – Drawing particulars from universals cautiously: Statements: All rats are bats. Some bats are desks. All desks are chairs. Conclusions: I) Some desks are rats. II) Some chairs are rats. Select the option that must be true.

Introduction / Context:
This problem mixes one universal and one particular with a second universal. The trap is to assume the particular bats that are desks must also be rats, or that chairs necessarily capture some rats. Neither is compelled by the premises.

Given Data / Assumptions:

• All Rats are Bats (R ⊆ B).
• Some Bats are Desks (∃ B ∩ Dk).
• All Desks are Chairs (Dk ⊆ Ch).

Concept / Approach:
Conclusion I requires R ∩ Dk ≠ ∅. But 'Some B are Dk' could hold entirely in the part of B disjoint from R; nothing requires those Desk-bats to be rats. Conclusion II requires Ch ∩ R ≠ ∅. We only know R ⊆ B and Dk ⊆ Ch; without a link R ↔ Dk, Chairs might have no rats.

Step-by-Step Solution:
Build a countermodel for I: Partition B into two disjoint parts: one equals R, the other equals Dk. Then R ∩ Dk = ∅ while 'Some B are Dk' holds. Thus I is not necessary.For II: Since Dk ⊆ Ch but rats need not be desks, it is possible that R is entirely outside Ch. Hence II is also not necessary.

Verification / Alternative check:
Add explicit cardinalities: let R = {r1}, B = {r1, b1}, Dk = {b1}, Ch = {b1, c2}. All premises hold; neither I nor II holds. This concrete model confirms the non-necessity of both conclusions.

Why Other Options Are Wrong:
Any option claiming one or both conclusions follow ignores that the particular need not hit the R subset and that Chairs can exist without containing any rats.

Common Pitfalls:
Mistaken belief that a particular statement about the superset B must overlap every subset of B, and careless propagation of membership through unrelated universals.

Final Answer:
Neither Conclusion I nor II follows.