Syllogism — determine which conclusions necessarily follow Statements: • All politicians are honest. • All honest are fair. Conclusions to evaluate: I. Some honest are politicians. II. No honest is politician. III. Some fair are politicians. IV. All fair are politicians.

Introduction / Context:
This problem tests categorical (syllogistic) reasoning with universal statements. You must decide which of the listed conclusions must be true in every situation that satisfies the given statements. The trap is to avoid adding information that is not guaranteed and to use subset logic properly.

Given Data / Assumptions:

• All politicians are honest. Denote sets: P = politicians, H = honest, F = fair.
• All honest are fair.
• Standard test convention assumes the named categories are nonempty unless explicitly ruled out.

Concept / Approach:
Translate the statements into set relations using subset notation. From the premises we have P ⊆ H and H ⊆ F, so by transitivity P ⊆ F. A conclusion of the form some X are Y follows when there is at least one element of X and the relation forces it to be in Y. Conclusions claiming no or all must be checked carefully to ensure the premises really force such exclusions or universal containments.

Step-by-Step Solution:

Verification / Alternative check:
Construct a concrete model. Let P = {p1}. Let H = {p1, h2}. Let F = {p1, h2, f3}. Then all politicians are honest and all honest are fair hold. I and III are true because p1 witnesses both. II and IV are false in this model, proving they are not necessary.

Why Other Options Are Wrong:

• None follows: false because I and III must hold under the usual nonemptiness convention.
• Only I follows: misses III, which is equally forced by P ⊆ F.
• Only I and II follow: includes II which contradicts the premises.
• None of these: option D already captures the correct pair.

Common Pitfalls:
Forgetting transitivity (P ⊆ H ⊆ F), assuming the converse F ⊆ P, or ignoring the existence of at least one politician.

Final Answer:
Only I and III follow