Statements: 1) All goats are cows. 2) All cows are animals. Conclusions: I. All goats are animals. II. All animals are goats. Choose the option that must follow.

Introduction / Context:
This problem tests two essentials: (i) chaining universal affirmatives (“All A are B,” “All B are C”) and (ii) spotting the converse fallacy (reversing a subset statement).

Given Data / Assumptions:

• Goats ⊆ Cows.
• Cows ⊆ Animals.

Concept / Approach:
Transitivity of subset: If A ⊆ B and B ⊆ C, then A ⊆ C. Conversely, from A ⊆ B we cannot infer B ⊆ A. The second conclusion is exactly that invalid reversal at a larger scale.

Step-by-Step Solution:
1) Chain the inclusions: Goats ⊆ Cows and Cows ⊆ Animals ⇒ Goats ⊆ Animals. So Conclusion I is guaranteed.2) Conclusion II (“All Animals are Goats”) would require Animals ⊆ Goats. That is a converse of an inclusion we do not have; it is not entailed.

Verification / Alternative check:
Model: Animals = {cow1, dog1}, Cows = {cow1}, Goats = {goat1} with goat1 = cow1. Premises hold, I is true (goat1 is an animal). However, dog1 is an animal but not a goat, falsifying II. Hence only I follows.

Why Other Options Are Wrong:
“Both” includes the converse fallacy; “Only II” is just the converse fallacy. “Neither” ignores the valid transitive implication.

Common Pitfalls:
Equating subset with equality. “All goats are cows” does not mean “only goats are cows,” and certainly not that “all animals are goats.”

Final Answer:
Only Conclusion I follows.