Introduction / Context:
The three premises create two chains into the set 'boxes': one via needles → threads → boxes and another via trees → boxes. The task is to see what must be true about potential overlaps.
Given Data / Assumptions:
- N ⊆ T (all needles are threads)
- T ⊆ B (all threads are boxes)
- Tr ⊆ B (all trees are boxes)
- Standard test convention assumes the subject class exists when drawing a particular conclusion (e.g., at least one needle exists).
Concept / Approach:
The two chains meet in B but the premises do not force an overlap between needles and trees. Hence, either they do not overlap at all, or they overlap in some elements. Also, since needles end up inside boxes, there is at least one box that is a needle if a needle exists.
Step-by-Step Solution:
III: Some boxes are needles. Because N ⊆ T ⊆ B and at least one needle exists, there is at least one box that is a needle. So III follows.I vs IV: The data do not settle whether any tree is a needle. It may be that no needle is a tree (I) or that some trees are needles (IV). Exactly one of these will be true in any model, but which one is not determined by the premises.II: Some trees are threads is not forced. Trees and threads are each subsets of boxes, but need not overlap.
Verification / Alternative check:
Scenario A: Place needles on one side of B and trees on the other (disjoint). Then I holds and IV fails. Scenario B: Let some needles also be trees. Then IV holds and I fails. III is true in both scenarios.
Why Other Options Are Wrong:
A: Incorrect because III definitely follows.B or C: These omit III or wrongly include II.D: Only III ignores the necessary either–or between I and IV.
Common Pitfalls:
Assuming sets inside a larger set must intersect; thinking that two subsets of the same superset automatically overlap.
Final Answer:
Only either I or IV, and III follow
Discussion & Comments