Difficulty: Easy
Correct Answer: Only I and II follow
Explanation:
Introduction / Context:
Use the universal statement “all beads are rings” to move any known bead elements into the ring set. Combine that with “some hammers are beads” to produce a guaranteed overlap between rings and hammers. Separately, the “some boxes are hammers” statement converts to “some hammers are boxes.”
Given Data / Assumptions:
Concept / Approach:
Push existence forward: elements in Bd are also in Rg. Therefore, the specific hammers that are beads are necessarily hammers that are rings. However, to claim “some rings are boxes,” we would need an element known to be both ring and box; the premises do not force that overlap.
Step-by-Step Solution:
I: Because some hammers are beads and all beads are rings, those elements are in Hm ∩ Rg. So “some rings are hammers” is true. I follows.II: From “some boxes are hammers,” we can rephrase equivalently as “some hammers are boxes.” II follows.III: “Some rings are boxes” would require a single element that is both a ring (via bead) and a box. The hammer that is a box may be different from the hammer that is a bead, so this need not hold. III does not follow.
Verification / Alternative check:
Build a model with disjoint subsets of hammers: one subset overlaps boxes; another subset overlaps beads (hence rings). All premises hold, but there is no ring–box overlap, confirming only I and II are guaranteed.
Why Other Options Are Wrong:
Common Pitfalls:
Assuming two separate “some” statements must refer to the same individuals; forgetting valid reversal of a “some” statement (A ∩ B implies B ∩ A).
Final Answer:
Only I and II follow
Discussion & Comments