Syllogism – Particulars with one universal inclusion: Statements: Some roads are ponds. All ponds are stores. Some stores are bags. Conclusions: I) Some bags are ponds. II) Some stores are roads. Choose the necessarily true conclusion(s).

Introduction / Context:
This item asks you to separate what is forced from what is merely possible when two particular statements are present along with a universal inclusion. The terms are Roads (Rd), Ponds (P), Stores (S), and Bags (B).

Given Data / Assumptions:

• Some Rd are P (∃ Rd ∩ P).
• All P are S (P ⊆ S).
• Some S are B (∃ S ∩ B).

Concept / Approach:
From the first two premises, the specific roads that are ponds are also stores. Therefore there exists at least one element in S that is a Road, guaranteeing Conclusion II. Conclusion I would require overlap between B and P, but 'Some S are B' could be satisfied by stores outside P.

Step-by-Step Solution:
Take an element x in Rd ∩ P (exists by premise). Because P ⊆ S, x is also in S. Therefore S ∩ Rd ≠ ∅, proving 'Some stores are roads' (II).For I, we would need B ∩ P ≠ ∅. Premise 3 gives only B ∩ S ≠ ∅, which might lie in S \ P. Hence I is not necessary.

Verification / Alternative check:
Countermodel for I: Let P be one subset of S associated with certain roads, and let B intersect S in a different subset disjoint from P. All premises remain true, but no bag is a pond. Therefore I fails while II holds.

Why Other Options Are Wrong:

• Only I follows / Both follow / Either follows: I is not compelled.
• Neither follows: incorrect since II is guaranteed by subset propagation.

Common Pitfalls:
Assuming that two 'some' statements about the same superset S must intersect the same region of S. They need not.

Final Answer:
Only Conclusion II follows.