Difficulty: Easy
Correct Answer: None of these
Explanation:
Introduction / Context:
A key rule in syllogisms is that universal statements like “all A are B” do not by themselves guarantee the existence of any A. Conclusions that use “some” require at least one actual instance. We must therefore separate the safe subset propagation from any unjustified existential claims.
Given Data / Assumptions:
Concept / Approach:
From C ⊆ B and B ⊆ W, we can derive C ⊆ W, i.e., “all cars are windows.” However, “some cars are dolls” and “some windows are cars” each require proof that at least one car exists. The premises never assert existence of cars; therefore such “some”-type conclusions are not logically forced.
Step-by-Step Solution:
I: All cars are windows: C ⊆ B and B ⊆ W imply C ⊆ W. The relationship holds universally but does not assert that any car actually exists.II: Some cars are dolls: This would need C ∩ D ≠ ∅. No statement links cars with dolls beyond both being inside windows, and existence of cars is not guaranteed. II does not follow.III: Some windows are cars: This needs C to be non-empty. Universals alone do not provide existence; thus III is not compelled.
Verification / Alternative check:
Consider a model with zero cars. All premises remain true, I remains a valid universal relation, but II and III fail due to lack of existence, proving they do not follow necessarily.
Why Other Options Are Wrong:
Common Pitfalls:
Assuming existential import from universal premises; inferring overlap merely because two classes share a larger containing class.
Final Answer:
None of these
Discussion & Comments