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.

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:

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.