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).
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ANone follows
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BOnly I follow
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COnly II follows
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DOnly III follows
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ENone of these
Answer
Correct Answer: Only I follow
Explanation
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:
I: Take y with y ∈ Sparrows ∩ Crows. Since Sparrows ⊆ Koels, y ∈ Koels as well ⇒ Koels ∩ Crows ≠ ∅. I follows.II: “Some parrots are crows” needs Parrots ∩ Crows ≠ ∅; there is no premise linking Parrots to Crows.III: “Some sparrows are pigeons” would require the Sparrow–Crow witnesses also to be Pigeons (or another link); not guaranteed.IV: “No crow is a parrot” is a strong universal negative absent from the premises; cannot be concluded.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.