Syllogism — Identify the valid conclusion(s): Statements: • All cycles are ducks. • All ducks are swans. Conclusions: I. All swans are cycles. II. Some swans are cycles.

Introduction / Context:
A classic chain of two universal affirmatives tests understanding of transitivity and the difference between a converse and an existential consequence.

Given Data / Assumptions:

• Cycles ⊆ Ducks.
• Ducks ⊆ Swans.

Concept / Approach:
From two subset relations, we get Cycles ⊆ Swans. However, the converse “All Swans are Cycles” is not implied. If there exists at least one Cycle (standard exam convention assumes non-empty classes in such everyday terms), then at least one Swan (that cycle) exists which is also a Cycle, yielding “Some Swans are Cycles.”

Step-by-Step Solution:
1) Chain: Cycles ⊆ Ducks ⊆ Swans ⇒ Cycles ⊆ Swans.2) C1 (“All swans are cycles”) would require Swans ⊆ Cycles; not given and generally false.3) C2 (“Some swans are cycles”) follows if Cycles is non-empty; any Cycle is automatically a Swan.

Verification / Alternative check:
Construct a diagram with a small Cycles set inside a larger Swans set; clearly, some Swans (those that are cycles) exist.

Why Other Options Are Wrong:
They either include C1 (the converse) or exclude the existential that follows from a non-empty Cycles set.

Common Pitfalls:
Treating “All A are B” as if it implied “All B are A,” and forgetting to check existence for “Some …” conclusions.

Final Answer:
Only Conclusion II follows.