Syllogism — Premises: All Americans are English-speaking. No Eskimos are English-speaking. Conclusions: I) No Eskimos are Americans. II) No English-speakers are Eskimos.

Difficulty: Easy

Correct Answer: Both I and II follow

Explanation:


Introduction / Context:
This is a classic combination of a universal inclusion and a universal exclusion. We test whether each conclusion is forced.


Given Data / Assumptions:

  • Americans ⊆ EnglishSpeaking.
  • Eskimos ∩ EnglishSpeaking = ∅ (equivalently, Eskimos ⊆ Not-EnglishSpeaking).


Concept / Approach:
If Eskimos share no members with English-speaking, and Americans are entirely within English-speaking, then Eskimos cannot be Americans. Also, 'No Eskimos are English-speaking' is equivalent to 'No English-speakers are Eskimos' in set-theoretic negation.


Step-by-Step Solution:

I) If Americans ⊆ EnglishSpeaking and Eskimos ∩ EnglishSpeaking = ∅, then Americans ∩ Eskimos = ∅. Hence no Eskimo is an American.II) The given exclusion is symmetric: A ∩ B = 0 implies B ∩ A = 0. So 'No English-speakers are Eskimos' follows directly.


Verification / Alternative check:
Venn diagrams show disjoint sets Eskimos and EnglishSpeaking, with Americans inside EnglishSpeaking; thus both conclusions hold.


Why Other Options Are Wrong:
Any option denying either I or II contradicts the disjointness and inclusion facts.


Common Pitfalls:
Forgetting that 'No A are B' is bidirectional in effect for membership claims (mutual exclusivity).


Final Answer:
Both I and II follow.

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