Two-step subset chain syllogism: 'All pens are roads' and 'All roads are houses' — test conclusions (i) All houses are pens, (ii) Some houses are pens, using inclusion logic and existence considerations

Given data

• Premise 1: All pens are roads (Pens ⊆ Roads).
• Premise 2: All roads are houses (Roads ⊆ Houses).
• Conclusions: (I) All houses are pens. (II) Some houses are pens.

Concept/Approach (why this method)

Chain the subsets and analyze direction. Universal affirmatives compose forward but do not reverse. Particular existence of Pens yields an existential conclusion inside Houses.

Step-by-Step calculation (logical derivation)
1) From the chain: Pens ⊆ Roads ⊆ Houses ⇒ Pens ⊆ Houses.2) Conclusion I requires Houses ⊆ Pens (reverse inclusion) — not implied ⇒ I is false.3) If at least one pen exists (standard test assumption), that pen is a house ⇒ 'Some houses are pens' is true ⇒ II follows.

Verification/Alternative

Venn placement: a small Pens circle inside Roads, itself inside Houses. The existence of any pen ensures a non-empty overlap (some Houses are Pens), but not the reverse containment.

Common pitfalls

• Assuming reverse inclusion (converse).
• Confusing universal statements with existential implications without noting existence of the subject class.

Final Answer
Only conclusion II follows.