Introduction / Context:
This is a classic chain with one particular premise. By following the chain carefully, you can see which subsets must overlap and which cannot.
Given Data / Assumptions:
- P ∩ C ≠ ∅ (some papers are cats)
- C ⊆ B (all cats are bats)
- B ∩ H = ∅ (no bat is a horse)
Concept / Approach:
Combine the particular statement with the subset inclusion: the particular papers that are cats must also be bats. With the universal negative regarding horses, you can infer impossibilities and necessary inclusions.
Step-by-Step Solution:
III: Some bats are papers. The papers that are cats are also bats, so at least one bat is a paper. Hence III follows.II: No horse is a cat. Since all cats are bats and no bat is a horse, cats and horses are disjoint. Therefore II follows.I: Some papers are horses. Not supported. The particular papers known to us (those which are cats) cannot be horses, and we do not know anything about other papers being horses.IV: All papers are bats. Overgeneralization. Only the papers that are cats must be bats; other papers need not be.
Verification / Alternative check:
Draw a Venn sketch: place a non-empty overlap of papers and cats within bats; keep horses disjoint from bats. Then II and III hold, I and IV do not.
Why Other Options Are Wrong:
A and D include I, which does not follow. C includes IV, which overstates the relationship. E claims all, which is incorrect.
Common Pitfalls:
Assuming that a property of some papers extends to all papers; ignoring that a universal negative about bats and horses blocks any cat–horse overlap.
Final Answer:
Only II and III follow
Discussion & Comments