Syllogism – Two universals in a chain: Statements: (a) All animals are dogs. (b) All dogs are birds. Conclusions: I) All animals are birds. II) All birds are animals.

Introduction / Context:A simple transitive chain of universals yields one necessary inclusion; the converse does not hold.

Given Data / Assumptions:

• Animals ⊆ Dogs ⊆ Birds.

Concept / Approach:Transitivity produces Animals ⊆ Birds (I). The reverse “All birds are animals” adds information not provided and is not necessary.

Step-by-Step Solution:From Animals ⊆ Dogs and Dogs ⊆ Birds, deduce Animals ⊆ Birds.

Verification / Alternative check:Countermodel for II: Let Birds include animals and non-animals (e.g., mechanical birds); universals (a) and (b) can still hold while II fails.

Why Other Options Are Wrong:They either include the invalid converse or deny the valid transitive inclusion.

Common Pitfalls:Assuming symmetry of subset relations without basis.

Final Answer:Only Conclusion I follows.