Syllogism – Particular plus universal, reasoning about non-barking dogs: Statements: • Some dogs bark. • All dogs bite. Conclusions: I) Dogs that do not bark also bite. II) Dogs that do not bark do not necessarily bite.

Introduction / Context:We are asked to reason about the subset of dogs that do not bark, given that all dogs bite.

Given Data / Assumptions:

• Some Dogs ⊆ Bark.
• Dogs ⊆ Bite (i.e., every dog bites).

Concept / Approach:Since “bite” applies to all dogs, it also applies to every sub-group of dogs, including the non-barking ones. Therefore (I) is necessarily true and (II) is false.

Step-by-Step Solution:Let NB = Dogs \ Bark (non-barking dogs). Because Dogs ⊆ Bite, NB ⊆ Bite as well.Hence every non-barking dog bites, proving (I).(II) contradicts this and is not necessary.

Verification / Alternative check:No model can satisfy “All dogs bite” and produce a non-biting dog, barking or not.

Why Other Options Are Wrong:They deny or weaken the universal inclusion.

Common Pitfalls:Taking “some dogs bark” to restrict “all dogs bite” to only barking dogs.

Final Answer:Only I follows.