Syllogism — mixing “some”, “no”, and a universal inclusion Statements: • Some towels are brushes. • No brush is soap. • All soaps are rats. Conclusions to evaluate: I. Some rats are brushes. II. No rat is brush. III. Some towels are soaps.

Introduction / Context:
This question blends an existential overlap with a universal negative and a universal inclusion. The aim is to see whether any necessary relationship can be forced between brushes and rats or between towels and soaps.

Given Data / Assumptions:

• T = towels, B = brushes, S = soaps, R = rats.
• Some T are B.
• No B is S.
• All S are R.

Concept / Approach:
No B is S tells us brushes and soaps are disjoint. All S are R only places soaps inside rats; it does not say all rats are soaps. Therefore, brushes might still be rats (if there exist rats outside S) or might not be; the premises are silent. Also, nothing connects towels to soaps except through brushes, and brushes are explicitly not soaps.

Step-by-Step Solution:

Verification / Alternative check:
Model A: Let S ⊆ R and R also contain r1 not in S. Let B = {r1} and some T include r1. Then I is true and II is false. Model B: Let B disjoint from R entirely, and T overlap B. Then II is true and I false. Since different valid models yield different truth values, none of I–III is necessary.

Why Other Options Are Wrong:

• Only either I or II follows: although exactly one of them will be true in any single model, the premises do not force which one; therefore the disjunction is not a necessary consequence here.
• Only II follows / Only I and III follow: each asserts claims that can fail under valid models.
• None of these: option A already captures that no listed conclusion must hold.

Common Pitfalls:
Assuming all rats are soaps (converse of all S are R); trying to pass through the brush set to reach soaps despite the explicit exclusion.

Final Answer:
None follows