Syllogism – Chains of inclusion across three sets Premises: 1) All branches are flowers. 2) All flowers are leaves. Conclusions to test: I) All branches are leaves. II) All leaves are branches. III) All flowers are branches. IV) Some leaves are branches.

Introduction / Context:
Syllogism questions with three terms often require chaining universal statements. Here, “All branches are flowers” and “All flowers are leaves” must be combined to check which conclusions necessarily hold. We also consider whether a particular (“some”) conclusion is justified given the premises and typical test conventions.

Given Data / Assumptions:

• B = set of branches.
• F = set of flowers.
• L = set of leaves.
• Premises: B ⊆ F and F ⊆ L.

Concept / Approach:
From subset chaining, B ⊆ L (transitivity). Claims reversing the subset (like “All leaves are branches”) are invalid. Whether we can state “Some leaves are branches” typically depends on whether B is non-empty. Most exam conventions assume the subject class exists, allowing a valid particular inference from the established inclusion.

Step-by-Step Solution:

Verification / Alternative check:

Why Other Options Are Wrong:

Common Pitfalls:

Final Answer:
Only I and IV follow