Difficulty: Medium
Correct Answer: Only conclusions I, II and III follow
Explanation:
Introduction / Context:We have a chain of three universal inclusions. We must test a mix of universal and existential conclusions.
Given Data / Assumptions:
Concept / Approach:By transitivity, Books ⊆ Papers (II). Therefore, if any book exists, it is a paper, validating I. Also, Books ⊆ Pens, so some pens are books (III) under non-emptiness. IV (All boxes are books) reverses inclusion and is invalid.
Step-by-Step Solution:
Step 1: Compose subsets: Books ⊆ Boxes ⊆ Pens ⊆ Papers.Step 2: Conclude II: All books are papers — true.Step 3: Existentials: pick any book b; since b ∈ Papers, I is true. Also b ∈ Pens, so III is true.Step 4: IV asserts Boxes ⊆ Books, which is converse and not supported.Verification / Alternative check:A concrete chain example quickly confirms I, II, III while falsifying IV by allowing boxes that are not books.
Why Other Options Are Wrong:Any option denying any of I–III or asserting IV conflicts with the chain.
Common Pitfalls:Overlooking that existential conclusions require at least one member, typically assumed in exam problems.
Final Answer:Only conclusions I, II and III follow.
Discussion & Comments