Syllogism — Peacocks, tigers, and lions (subset chaining): Statements: (a) All peacocks are lions. (b) Some tigers are peacocks. Conclusions to test: I. Some lions are not tigers. II. All tigers are lions. III. Some tigers are lions. IV. All peacocks are tigers. Choose the option that states what necessarily follows.

Introduction / Context:
This problem checks your command over categorical syllogisms with subset chaining and the difference between “some” and “all.” We must mark only the conclusions that are logically necessary under every model that satisfies the statements.

Given Data / Assumptions:

• (a) All peacocks are lions (Peacock ⊆ Lion).
• (b) Some tigers are peacocks (∃ Tiger ∩ Peacock).
• Standard test convention: existence is guaranteed only where explicitly stated (“some”).

Concept / Approach:
Use set inclusion and intersection. If some tigers are peacocks and every peacock is a lion, those specific tigers are lions. No other universal or negative relation is licensed by the premises.

Step-by-Step Solution:
1) From (b), pick at least one entity that is both Tiger and Peacock.2) From (a), Peacock ⊆ Lion, so that entity is also a Lion.3) Therefore, ∃ (Tiger ∩ Lion): “Some tigers are lions” (III) must be true.4) I “Some lions are not tigers” is not forced; all lions could, in principle, be tigers.5) II “All tigers are lions” generalizes far beyond the premises and is not warranted.6) IV “All peacocks are tigers” reverses the inclusion direction and is unsupported.

Verification / Alternative check:
Create a model: Peacocks form a small subset inside Lions; some Tigers overlap that Peacock subset. III is compelled; I, II, and IV can vary without violating the premises.

Why Other Options Are Wrong:
Each of I, II, IV adds information not entailed by (a) and (b).

Common Pitfalls:
Assuming “some” implies “all,” or reversing subset directions.

Final Answer:
Only conclusion III follows.