Logical syllogism — determine which conclusions follow beyond doubt. Statements: 1) All the books are pencils. 2) No pencil is an eraser. Conclusions: (1) All the pencils are books. (2) Some erasers are books. (3) No book is an eraser. (4) Some books are erasers.

Introduction / Context:
This syllogism mixes a universal affirmative with a universal negative. We need to deduce what must be true about the relationship between books and erasers.

Given Data / Assumptions:

• All books ⊆ pencils.
• No pencil is an eraser.

Concept / Approach:
If a set A is a subset of B, and B is disjoint from C, then A is also disjoint from C. Hence, from “All books are pencils” and “No pencil is an eraser,” we get “No book is an eraser.”

Step-by-Step Solution:
(3) “No book is an eraser” — follows directly from subset plus disjointness. (1) “All pencils are books” — converse of the first premise; not implied. (2) “Some erasers are books” — contradicts (3); cannot follow. (4) “Some books are erasers” — contradicts (3); cannot follow.

Verification / Alternative check:
Draw pencils and erasers as disjoint sets; place books entirely within pencils. There is no overlap between books and erasers, proving (3) only.

Why Other Options Are Wrong:
Options including (1) assume an illegitimate converse; options including (2) or (4) contradict the necessary disjointness.

Common Pitfalls:
Overlooking that a subset of a set disjoint from C must also be disjoint from C.

Final Answer:
Only (3)