Difficulty: Medium
Correct Answer: Both III & IV follow
Explanation:
Introduction / Context:Two chains appear here: one establishing that bananas sit entirely inside guavas, and another that places some apples inside bananas. We must see what this forces about guavas and apples/bananas, while handling grapes and pomegranates cautiously.
Given Data / Assumptions:
Concept / Approach:From y ∈ Apples ∩ Bananas and Bananas ⊆ Guavas, the same y is a Guava and an Apple, proving “Some guavas are apples.” Since every Banana is a Guava and no Guava is a Pomegranate, it follows that no Banana is a Pomegranate. However, statements involving Grapes require the same Apple witness to be a Grape, which is not forced.
Step-by-Step Solution:
III: Take y with y ∈ Apples ∩ Bananas. Because Bananas ⊆ Guavas, y ∈ Guavas; thus Guavas ∩ Apples ≠ ∅ ⇒ III follows.IV: From Bananas ⊆ Guavas and Guavas ∩ Pomegranates = ∅, we get Bananas ∩ Pomegranates = ∅ ⇒ IV follows.I: “No grapes are pomegranates” is too strong; the Grapes–Apples overlap does not force those grapes to be bananas/guavas.II: “Some guavas are grapes” would need the Apple–Banana witness y also to be a Grape; not guaranteed.Verification / Alternative check:Model where some apples are bananas (hence guavas), and independently some grapes are apples but not the banana apples. Then III and IV hold; I and II fail.
Why Other Options Are Wrong:They include conclusions about grapes that depend on an overlap not enforced by the premises.
Common Pitfalls:Assuming that two separate “some” statements over Apples refer to the same individuals.
Final Answer:Both III & IV follow.
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