Syllogism – Universal inclusion with a universal exclusion: Statements: (a) All pens are pencils. (b) No pencil is a monkey. Conclusions: I) No pen is a monkey. II) Some pens are monkeys. III) All monkeys are pens. IV) Some monkeys are pens.

Introduction / Context:A chain of subset then disjointness produces a clean negative conclusion about Pens and Monkeys.

Given Data / Assumptions:

• Pens ⊆ Pencils.
• Pencils ∩ Monkeys = ∅.

Concept / Approach:If Pens sit inside Pencils, and Pencils are disjoint from Monkeys, then Pens and Monkeys are disjoint. That proves I. Claims II, III, IV contradict or go beyond this disjointness.

Step-by-Step Solution:From Pens ⊆ Pencils and Pencils ∩ Monkeys = ∅, deduce Pens ∩ Monkeys = ∅.Therefore “No pen is a monkey” (I) is necessarily true.

Verification / Alternative check:No countermodel can make a pen also a monkey while keeping “No pencil is a monkey” and “All pens are pencils”.

Why Other Options Are Wrong:They contradict the disjointness or claim arbitrary either/or without support.

Common Pitfalls:Missing the propagation of exclusion down the subset chain.

Final Answer:Only conclusion I follows.