Syllogism – Chain reasoning with “some” and a universal: Statements: 1) Some spoons are cups. 2) Some cups are bottles. 3) All bottles are mugs. Conclusions: I) Some mugs are cups. II) Some cups are bottles. III) Some spoons are mugs. Select the correct set of conclusions that necessarily follow.

Introduction / Context:
This test mixes two particular premises with one universal. We must determine which conclusions are guaranteed across all interpretations.

Given Data / Assumptions:

• Some Spoons are Cups (∃ S∩C).
• Some Cups are Bottles (∃ C∩B).
• All Bottles are Mugs (B ⊆ M).

Concept / Approach:
Push “some” through a subset only when the element is known to be in that subset; be careful with intersections that could be disjoint.

Step-by-Step Solution:
For II: “Some cups are bottles.” This is exactly a premise; therefore it necessarily follows.For I: From “Some cups are bottles” and “All bottles are mugs,” those specific cups are also mugs. Hence there exist elements that are both Cups and Mugs: “Some mugs are cups.”For III: “Some spoons are mugs.” We know some S are C, and some (possibly different) C are B ⊆ M. But the spoon-cups might be disjoint from the cup-bottles. No forced overlap, so III does not necessarily follow.

Verification / Alternative check:
Construct a countermodel for III: let S∩C be non-bottle cups; let C∩B be separate cups. Then no spoon is a mug, falsifying III while keeping premises true.

Why Other Options Are Wrong:

• Only I follows: misses II which is directly given.
• All follow: III is not necessary.
• Only II follows: I is also forced via subset propagation.
• None of these: Incorrect because “Only I and II follow” is correct.

Common Pitfalls:
Assuming that two “some” statements about the same middle term (cups) must intersect; they need not.

Final Answer:
Only I and II follow.