Syllogism – Two chained universals (subset of subset): Statements: All Asians are wise. All Chinese are Asian. Therefore, choose the necessary conclusion.

Introduction / Context:
This is a textbook example of subset chaining in syllogisms. If set C (Chinese) is a subset of set A (Asian), and set A is a subset of set W (Wise), then C is a subset of W. The task is to select the conclusion that is true in every possible model satisfying the premises.

Given Data / Assumptions:

• All Chinese are Asian (C ⊆ A).
• All Asians are Wise (A ⊆ W).
• No 'some' premise is provided; we cannot infer existence of a Chinese individual from universals alone.

Concept / Approach:
Chaining subset relations is transitive: if C ⊆ A and A ⊆ W, then C ⊆ W. Statements about 'some' (existence) should not be inferred unless a particular premise guarantees it. That is why 'Some Chinese are wise' is not logically necessary under the most conservative interpretation.

Step-by-Step Solution:
From C ⊆ A and A ⊆ W, derive C ⊆ W by transitivity.The statement 'All Chinese are wise' exactly captures C ⊆ W and therefore must be true in every model respecting the premises.The statement 'Some Chinese are wise' would additionally require that C is non-empty, which the premises do not assert.

Verification / Alternative check:
Create a model with C empty, A and W non-empty. All universals remain true, and 'Some Chinese are wise' becomes false. This shows that 'some' does not necessarily follow, while 'All Chinese are wise' remains valid in all models.

Why Other Options Are Wrong:

• Some Chinese are wise: not guaranteed without an existence premise.
• All wise are Chinese: not supported; W may include non-Chinese.
• No conclusion follows: incorrect, as the transitive universal conclusion holds.

Common Pitfalls:
Confusing universal inclusion with existential import. Many learners try to infer 'some' from 'all', which is invalid in strict syllogistic logic used in aptitude tests.

Final Answer:
All Chinese are wise.