Syllogism — One existential plus a two-step inclusion chain Statements: • Some roses are flowers. • Some flowers are buds. • All buds are leaves. • All leaves are plants. Conclusions: I) Some plants are flowers. II) Some roses are buds. III) No leaves are roses. IV) No roses are buds. Select what must be true.
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AOnly I follows
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BOnly I & II follows
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COnly I and either II or IV follows
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DOnly either II and IV follows
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ENone of these
Answer
Correct Answer: Only I follows
Explanation
Introduction / Context:The structure provides a single existential inside Flowers (at Buds) and pushes it forward through two universal inclusions to Plants. The question is which conclusions are forced without assuming unintended overlaps.
Given Data / Assumptions:
- ∃b: b ∈ Flowers ∩ Buds.
- Buds ⊆ Leaves.
- Leaves ⊆ Plants.
- ∃r: r ∈ Roses ∩ Flowers (separate existential).
Concept / Approach:From b ∈ Flowers and Buds ⊆ Leaves ⊆ Plants, b is simultaneously a Flower and a Plant. Hence “Some plants are flowers” is guaranteed. However, nothing links Roses to Buds or to Leaves beyond Roses ⊆? None. Therefore, claims about Roses∩Buds or universal negatives involving Roses do not follow.
Step-by-Step Solution:
I: Take b ∈ Flowers ∩ Buds. Since Buds ⊆ Leaves ⊆ Plants, b ∈ Plants. Thus Plants ∩ Flowers ≠ ∅ ⇒ I follows.II: “Some roses are buds” would need overlap Roses ∩ Buds; no premise forces it.III & IV: Universal negatives about Leaves and Roses or Roses and Buds are not supported and may be false in many models.Verification / Alternative check:Create a model with a bud that is a flower and hence a plant, and with roses that are flowers but not buds. Only I is necessary.
Why Other Options Are Wrong:They include claims that require extra intersections or unjustified universal negatives.
Common Pitfalls:Assuming that two “some” statements within the same larger set must overlap; they need not.
Final Answer:Only I follows.