Introduction / Context:
This problem combines a universal negative with two universal affirmatives. The key is to trace each conclusion back to the premises without importing assumptions that are not warranted.
Given Data / Assumptions:
- H ∩ S = ∅ (no house is a school)
- C ⊆ S (all colleges are schools)
- S ⊆ T (all schools are teachers)
Concept / Approach:
Use transitivity for subset chains and apply the disjointness of houses and schools to specific subsets of S. Be careful with I: saying no house is a teacher would require that all teachers are schools, which is not given.
Step-by-Step Solution:
II: All colleges are teachers. Since C ⊆ S and S ⊆ T, we have C ⊆ T. Therefore II follows.IV: No college is a house. Because C ⊆ S and H ∩ S = ∅, the intersection C ∩ H is empty. Therefore IV follows.III: Some teachers are not houses. At least one school exists in typical test contexts (e.g., there is at least one college or school). Any school is a teacher and not a house (due to H ∩ S = ∅). Hence there exists at least one teacher that is not a house. Therefore III follows.I: No house is a teacher. This does not follow. Teachers may include members outside schools; such teachers could, in principle, be houses. The premises do not forbid that.
Verification / Alternative check:
Construct a model with teachers beyond schools (e.g., tutors not in S). Some of those could be houses without violating the premises, showing I is not necessary.
Why Other Options Are Wrong:
A: Rejects true conclusions II, III, IV.B and E: Use an either–or involving I, but I does not follow at all.D: Includes I, which is not guaranteed.
Common Pitfalls:
Assuming the converse of a subset relation (that all teachers must be schools); overextending a universal negative to a larger superset.
Final Answer:
Only II, III and IV follow
Discussion & Comments