Statements: • All papers are bottles. • All bottles are cups. • Some cups are jugs. • Some jugs are plates. Conclusions: I. Some plates are cups. II. Some plates are bottles. III. Some cups are papers. IV. Some bottles are papers. Choose the option that must follow.

Introduction / Context:
Two universal inclusions are followed by two existential claims about later sets. We must be careful not to assert intersections that depend on the very same jug being both a cup and a plate.

Given Data / Assumptions:

• Papers ⊆ Bottles ⊆ Cups.
• ∃c_j ∈ Cups ∩ Jugs.
• ∃j_p ∈ Jugs ∩ Plates.

Concept / Approach:
I and II require that the Cup-Jug element c_j is also a Plate (or the Jug-Plate element j_p is also a Cup). The premises do not force that identity. III and IV require existence of Papers; universal statements do not provide an existential witness.

Step-by-Step Solution:
• I–II: Not guaranteed because c_j and j_p can differ.• III–IV: “Some X are Papers” needs Papers to exist; not given.

Verification / Alternative check:
Let Papers be empty; let c_j and j_p be distinct. All premises remain true and I–IV fail.

Why Other Options Are Wrong:
They assume identity of witnesses or existential import not supplied.

Common Pitfalls:
Creating unintended chains across separate “some” statements.

Final Answer:
None follows.