Syllogism (Categorical Logic) — Cups, Glasses, Bowls, Plates Statements: • All cups are glasses. • Some glasses are bowls. • No bowl is a plate. Conclusions: I. No cup is a plate. II. No glass is a plate. III. Some plates are bowls. IV. Some cups are not glasses.

Introduction / Context:
This problem asks which conclusions must be true, given relationships among four sets (cups, glasses, bowls, plates). The trick is to avoid overreaching from “some” information and to respect that “No bowl is a plate” restricts only bowls, not all glasses or all cups.

Given Data / Assumptions:

• All cups ⊆ glasses.
• Some glasses ∩ bowls ≠ ∅.
• Bowls ∩ plates = ∅.

Concept / Approach:
Check each conclusion against only what is guaranteed by the premises. Do not assume existence beyond what “some” asserts, nor add unstated disjointness. Remember that “some” does not imply “all,” and “no bowl is a plate” does not imply “no glass is a plate.”

Step-by-Step Solution:

Verification / Alternative check:
Create a model: Take a cup that is a glass but not a bowl, and permit some other glasses to be plates. This satisfies all premises while falsifying I and II as universal negatives. III directly contradicts a premise, and IV contradicts the universal inclusion.

Why Other Options Are Wrong:

• Any combination listing I–IV is unjustified; none of the four is compelled by the premises.

Common Pitfalls:
Assuming all glasses are bowls; conflating “No bowl is a plate” with “No glass is a plate”; overlooking that “some” applies only to a subset.

Final Answer:
None follows