Difficulty: Easy
Correct Answer: Only II follows
Explanation:
Introduction / Context:
Universal exclusion (no hammer is a ring) coexists with a universal inclusion (doors ⊆ windows) and an existential about rings and doors. We look for the conclusion that is compelled regardless of how other sets relate.
Given Data / Assumptions:
Concept / Approach:
From r_d ∈ Doors and Doors ⊆ Windows, r_d ∈ Windows, so Windows ∩ Rings ≠ ∅ (II true). I would need Stones to reach Windows via Doors or another link; not given. III (“No window is a stone”) is not implied, because nothing forbids some Window from being a Hammer (and thus possibly a Stone) apart from ring-related exclusions. IV contradicts the exclusion because Stones ⊆ Hammers and Hammers ∩ Rings = ∅.
Step-by-Step Solution:
• Push r_d through Doors ⊆ Windows to prove II.• Note that there is no route forcing Stones into Windows; I does not follow.• III is a universal negative about Windows and Stones with no supporting premise.• IV is impossible due to Hammers ∩ Rings = ∅ and Stones ⊆ Hammers.
Verification / Alternative check:
Model Windows as a large set containing Doors plus other items; allow some Windows to overlap Hammers without being Rings. Premises remain consistent; only II is guaranteed.
Why Other Options Are Wrong:
They add claims not enforced by the premises or contradict them.
Common Pitfalls:
Reading ring-related exclusions as forbidding all other overlaps with Windows.
Final Answer:
Only II follows.
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