Introduction / Context:
We are given two universal negatives and one universal affirmative. The task is to infer which relationships must hold involving the subset 'erasers'.
Given Data / Assumptions:
- Pa ∩ Pe = ∅ (no paper is a pen)
- Pe ∩ Pn = ∅ (no pen is a pencil)
- E ⊆ Pa (all erasers are papers)
Concept / Approach:
From E ⊆ Pa and Pa ∩ Pe = ∅, erasers cannot be pens. However, nothing here links pencils to papers or erasers, so be careful not to chain two disjointness statements in an invalid way (disjointness is not transitive).
Step-by-Step Solution:
III: No pen is an eraser. True, because any eraser is a paper, and no paper is a pen. Thus E ∩ Pe = ∅.II: No pencil is an eraser. Not forced. We have no premise connecting pencils to papers or erasers; 'no pen is a pencil' does not help.I: Some papers are erasers. This would require existence of at least one eraser. The premises do not guarantee that; some exams avoid asserting existence from 'all' statements. Hence I is not necessary.IV: All papers are erasers. Reverses the given subset; not supported.
Verification / Alternative check:
Construct a model with zero erasers. Then I is false, II is undetermined, III is trivially true, and IV is false. None of the option groupings matches this combination, so the meta-choice 'None of these' is correct.
Why Other Options Are Wrong:
A, B, C, D include II and/or I as necessary, which they are not, or claim all are true, which is incorrect.
Common Pitfalls:
Treating disjointness as transitive; presuming existence from a universal statement without support.
Final Answer:
None of these
Discussion & Comments