Syllogism with a subset chain — identify guaranteed existentials Statements: • Some bottles are drinks. • All drinks are cups. Conclusions to evaluate: I. Some bottles are cups. II. Some cups are drinks. III. All drinks are bottles. IV. All cups are drinks.

Introduction / Context:
This syllogism combines an existential premise with a universal subset relation. The challenge is to convert these into solid conclusions without committing illicit conversion errors.

Given Data / Assumptions:

• Bo = bottles, Dk = drinks, Cu = cups.
• Some Bo are Dk.
• All Dk are Cu (Dk ⊆ Cu).

Concept / Approach:
When some Bo are Dk and all Dk are Cu, the particular members that are drinks are necessarily cups. Also, because there exists at least one drink (witnessed by the bottle-drink element), we can assert some cups are drinks. Universal statements about all cups or all drinks being other classes are not supported.

Step-by-Step Solution:

Verification / Alternative check:
Model: Let Dk = {d1, d2}, Cu = {d1, d2, c3}, Bo = {b1 = d1}. Premises hold. I and II are true; III and IV are false in this model. Hence only I and II are necessary.

Why Other Options Are Wrong:

• Only II and III / Only II and IV / Only III and IV / Only I and IV: each includes at least one unsupported universal conclusion.

Common Pitfalls:
Illicit conversion of all Dk are Cu into all Cu are Dk; forgetting to exploit the existence granted by some Bo are Dk.

Final Answer:
Only I and II follow