Syllogism — Determine the logically necessary conclusion(s): Statements: • All fruits are leaves. • Some fruits are grapes. Conclusions: I. Some leaves are grapes. II. All grapes are fruits.

Introduction / Context:
This problem mixes a universal inclusion with a particular existence statement. The key is to transfer membership through a superset using “all,” while avoiding illicit conversion in the second conclusion.

Given Data / Assumptions:

• S1: Fruits ⊆ Leaves.
• S2: Some Fruits are Grapes (Fruits ∩ Grapes ≠ ∅).
• Conclusions: C1 “Some Leaves are Grapes.” C2 “All Grapes are Fruits.”

Concept / Approach:
When X ⊆ Y and Some X are Z, those “some X” are simultaneously in Y, yielding Some Y are Z. However, “Some Fruits are Grapes” does not imply “All Grapes are Fruits”; that would be an illicit converse.

Step-by-Step Solution:
1) From S2, pick an element x with Fruit(x) and Grape(x).2) From S1, Fruit(x) ⇒ Leaf(x).3) Therefore Leaf(x) and Grape(x), proving C1: Some Leaves are Grapes.4) C2 claims Grapes ⊆ Fruits, which is not given. Grapes might include non-fruits in some models; hence C2 does not follow.

Verification / Alternative check:
Venn sketch: Fruits entirely inside Leaves; Grapes overlapping Fruits partly. The overlap witnesses C1. But Grapes may also extend outside Fruits, so C2 is not necessary.

Why Other Options Are Wrong:
Options including C2 assume a converse of S2; “Neither” ignores the guaranteed overlap for C1.

Common Pitfalls:
Assuming that because some Fruits are Grapes, therefore all Grapes are Fruits. “Some” never licenses “all.”

Final Answer:
Only Conclusion I follows.