Syllogism – Buses, trains, rooms, and boats (universal negatives) Statements: • All trains are buses. • No room is a bus. • All boats are rooms. Conclusions to test: I. No boat is a train. II. No bus is a boat. III. No train is a room.

Introduction / Context:
Here we combine two universal statements (“all” and “no”) to derive universal negatives. When a class is fully inside another (all trains ⊆ buses) and that other class is disjoint from a third (no rooms are buses), then the inside class is also disjoint from the third (no trains are rooms).

Given Data / Assumptions:

• T ⊆ B (all trains are buses).
• R ∩ B = ∅ (no room is a bus).
• All boats are rooms (Bo ⊆ R).

Concept / Approach:
Use contrapositive-style reasoning with set relations. If class A is disjoint from B, then any subset of B is also disjoint from A. Also, if all X belong to a class disjoint from Y, then X and Y are disjoint.

Step-by-Step Solution:
III: From T ⊆ B and R ∩ B = ∅, it follows that T ∩ R = ∅, so “No train is a room.” III follows.I: Since Bo ⊆ R and R is disjoint from B, Bo is also disjoint from B. Because T ⊆ B, Bo ∩ T = ∅, hence “No boat is a train.” I follows.II: From Bo ⊆ R and R ∩ B = ∅, we have Bo ∩ B = ∅, so “No bus is a boat.” II follows.

Verification / Alternative check:
Draw three non-overlapping regions for buses and rooms, with trains placed entirely inside buses and boats entirely inside rooms. Visual inspection confirms all three conclusions.

Why Other Options Are Wrong:

• Any option omitting one of the three conclusions contradicts the clean set-separation logic.
• “None follows” is clearly false because all three are mandated by the premises.

Common Pitfalls:
Forgetting that a subset of a set disjoint from another is also disjoint from that other; misreading “no A is B” as “some A are not B.”

Final Answer:
All follow