Syllogism — Chained universals with an existential corollary Statements: 1) All elephants are men. 2) All men are socks. Conclusions: I) Some socks are elephants. II) All elephants are socks. Select the necessary conclusion(s).
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AOnly Conclusion I follows
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BOnly Conclusion II follows
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CBoth Conclusions I and II follow
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DNeither Conclusion I nor II follows
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ENone of these
Answer
Correct Answer: Both Conclusions I and II follow
Explanation
Introduction / Context:This problem uses a chain of two universal inclusions to test both a universal consequence and a standard existential corollary often assumed in exam-style syllogisms when categories are non-empty.
Given Data / Assumptions:
- Elephants ⊆ Men.
- Men ⊆ Socks.
- Ordinary-language assumption: the class “Elephants” is non-empty, permitting an existential conclusion.
Concept / Approach:Transitivity yields Elephants ⊆ Socks, which directly validates the universal conclusion (II). If at least one elephant exists, that elephant is a sock under the given premises, making “Some socks are elephants” true as well (I). Although the real-world content is absurd, syllogistic validity depends on form, not plausibility.
Step-by-Step Solution:
Step 1: Compose inclusions: Elephants ⊆ Men and Men ⊆ Socks ⇒ Elephants ⊆ Socks (validates II).Step 2: Pick one elephant e (non-emptiness). Since e ∈ Elephants ⊆ Socks, e ∈ Socks; therefore, some socks are elephants (validates I).Verification / Alternative check:Venn diagram with nested sets (Socks largest, Men inside Socks, Elephants inside Men) shows II immediately and makes I true when Elephants is non-empty — as typically intended in such problems.
Why Other Options Are Wrong:Any option omitting II ignores transitivity; omitting I ignores the routine existential corollary used in these exams.
Common Pitfalls:Rejecting a logically valid consequence because the content sounds unrealistic; in syllogisms, form rules over real-world semantics.
Final Answer:Both Conclusions I and II follow.