Syllogism — Determine which conclusions necessarily follow: Statements: • All windows are doors. • All buildings are doors. • All doors are boats. Conclusions: I. All windows are boats. II. All buildings are boats. III. Some boats are doors.

Introduction / Context:
A classic transitivity problem where two classes (windows, buildings) both sit inside “doors,” and doors sit inside “boats.” We must test three conclusions.

Given Data / Assumptions:

• Windows ⊆ Doors.
• Buildings ⊆ Doors.
• Doors ⊆ Boats.

Concept / Approach:
For universal statements, subset transitivity applies: if X ⊆ Y and Y ⊆ Z, then X ⊆ Z.

Step-by-Step Solution:
1) I: Windows ⊆ Doors and Doors ⊆ Boats ⇒ Windows ⊆ Boats. True.2) II: Buildings ⊆ Doors and Doors ⊆ Boats ⇒ Buildings ⊆ Boats. True.3) III: Since every door is a boat, provided doors exist, at least “some boats are doors.” In standard exam convention with universal chains like these, this is accepted as true.

Verification / Alternative check:
Diagram three nested sets: Windows and Buildings inside Doors, Doors inside Boats; the results are immediate.

Why Other Options Are Wrong:
Any option omitting one of I/II/III underestimates the force of the universal chain.

Common Pitfalls:
Overlooking that if Doors are Boats, then every existing Door is a Boat, which licenses the particular statement “Some boats are doors” under typical test assumptions.

Final Answer:
All follow.