Syllogism – Three chained universals: Statements: (a) All dogs are rats. (b) All rats are crows. (c) All crows are parrots. Conclusions: (I) All dogs are parrots. (II) Some parrots are dogs. (III) Some crows are dogs. (IV) All rats are dogs.

Introduction / Context:
We have a pure chain of universals creating nested subsets: Dogs ⊆ Rats ⊆ Crows ⊆ Parrots. We must see which conclusions are necessary under standard aptitude conventions (which typically assume the subordinate sets are non-empty).

Given Data / Assumptions:

• D ⊆ R ⊆ C ⊆ P.
• Standard test convention: the named sets are considered to have members unless negated.

Concept / Approach:
(I) follows directly from transitivity: D ⊆ P. (II) and (III) are particular statements that hold under the conventional assumption that D is non-empty (then some P are D; some C are D). (IV) reverses the first link and does not follow.

Step-by-Step Solution:
Chain the inclusions to get D ⊆ P (I).Because D is non-empty in typical exam readings, pick x ∈ D; then x ∈ P and x ∈ C, giving (II) and (III).(IV) “All rats are dogs” is the converse of (a) and is invalid.

Verification / Alternative check:
Diagram the nesting P ⊇ C ⊇ R ⊇ D. Choosing any dog yields a crow and a parrot simultaneously, validating (II) and (III) under the non-empty convention.

Why Other Options Are Wrong:
They either omit a valid consequence or include the false converse (IV).

Common Pitfalls:
Illicit converse; forgetting typical test convention about existence for named categories.

Final Answer:
Only (I), (II) and (III) follow.