Syllogism — Analyze conclusions from chained inclusions Statements: • All crows are birds. • All peacocks are crows. Conclusions: I) All peacocks are birds. II) All birds are peacocks. Choose which conclusion(s) logically follow(s) from the statements.

Introduction / Context:
This syllogism tests whether you can correctly propagate set inclusions through a chain and avoid the common mistake of reversing them. We are told that the class of crows lies wholly inside the class of birds, and the class of peacocks lies wholly inside the class of crows. We must check two conclusions about peacocks and birds.

Given Data / Assumptions:

• All crows are birds (Crows ⊆ Birds).
• All peacocks are crows (Peacocks ⊆ Crows).
• Standard reasoning-test assumption that these classes are meaningful (non-empty) in ordinary language, though non-emptiness is not required for the universal conclusion we draw.

Concept / Approach:
Subset relations compose transitively: if A ⊆ B and B ⊆ C, then A ⊆ C. However, the converse (e.g., C ⊆ A) cannot be inferred unless explicitly stated. Therefore, we can validly push inclusion forward but not backward.

Step-by-Step Solution:

Verification / Alternative check:
Draw three nested circles: Birds (largest), Crows inside Birds, and Peacocks inside Crows. You can see immediately I is true while II would force the largest set (Birds) to shrink to Peacocks, contradicting the diagram unless additional premises are given.

Why Other Options Are Wrong:

• Both conclusions: invalid because II reverses inclusion.
• Neither conclusion: incorrect because I is certain.
• Only conclusion II: incorrect — converse error.

Common Pitfalls:
Reversing 'All A are B' into 'All B are A' (the converse) and assuming equality of classes when only one-way inclusion is stated.

Final Answer:
Only conclusion I follows.