Syllogism – Chained universals with one particular: Statements: Some cats are dogs. All dogs are goats. All goats are walls. Conclusions: I) Some walls are dogs. II) Some walls are cats. Identify what necessarily follows.

Introduction / Context:
This problem examines how a single particular statement propagates through a chain of universal subset relations. The terms are Cats (C), Dogs (D), Goats (G), and Walls (W).

Given Data / Assumptions:

• Some C are D (∃ C ∩ D).
• All D are G (D ⊆ G).
• All G are W (G ⊆ W).

Concept / Approach:
A member that lies in D will also lie in G, and therefore in W, by successive subset inclusions. The particular witness provided by 'Some cats are dogs' is the seed that carries through the chain.

Step-by-Step Solution:
Pick an element x that is both a Cat and a Dog (from the 'some' premise).Because all Dogs are Goats, x is a Goat.Because all Goats are Walls, x is a Wall.Therefore x is a Dog and a Wall, proving 'Some Walls are Dogs' (Conclusion I).The same x is a Cat and a Wall, proving 'Some Walls are Cats' (Conclusion II).

Verification / Alternative check:
A simple three-circle Venn diagram with D inside G inside W and C overlapping D demonstrates a non-empty overlap between W and both D and C at the same point x.

Why Other Options Are Wrong:

• Only I or Only II: too weak; the single witness x verifies both statements.
• Neither / Either: contradicted by the forced overlaps via the universal chain.

Common Pitfalls:
Forgetting that a single 'some' element can satisfy multiple conclusions once universals propagate membership through nested sets.

Final Answer:
Both Conclusions I and II follow.