Syllogism — Propagate a specific witness through a universal inclusion Statements: • All parrots are pigeons. • Some crows are pigeons. • Some sparrows are crows. • All sparrows are koels. Conclusions: I) Some koels are crows. II) Some parrots are crows. III) Some sparrows are pigeons. IV) No crow is a parrot. Select the necessary conclusion(s).

Introduction / Context:
We have two separate “some” facts and two universal statements. The key is to identify where a particular witness can be pushed through a universal inclusion to force an intersection.

Given Data / Assumptions:

• Parrots ⊆ Pigeons.
• ∃x: x ∈ Crows ∩ Pigeons.
• ∃y: y ∈ Sparrows ∩ Crows.
• Sparrows ⊆ Koels.

Concept / Approach:
If some Sparrows are Crows and all Sparrows are Koels, then those very same individuals are both Crows and Koels, which proves “Some koels are crows.” The other conclusions either require connecting unrelated subsets or assert a universal negative without support.

Step-by-Step Solution:

Verification / Alternative check:
Model with Parrots and Pigeons disjoint from Crows, with a few Sparrows that are Crows (hence Koels), and some Crows that are Pigeons. Only I is forced.

Why Other Options Are Wrong:
They rely on intersections or universal negatives not supported by the given statements.

Common Pitfalls:
Assuming that multiple “some” facts must overlap in the same individuals; they need not.

Final Answer:
Only I follow.