Introduction / Context:
This syllogism mixes universal affirmatives and negatives. We must identify conclusions that must be true in every valid interpretation of the premises and recognize mutually exclusive pairings where exactly one of two alternatives can hold.
Given Data / Assumptions:
- All Trees are Flowers (T ⊆ F).
- No Flower is a Fruit (F ∩ Fr = ∅).
- All Branches are Fruits (B ⊆ Fr).
Concept / Approach:
- From T ⊆ F and F disjoint from Fr, we get T ∩ Fr = ∅, i.e., no Fruit is a Tree (equivalently, no Tree is a Fruit).
- From B ⊆ Fr and T ∩ Fr = ∅, Trees and Branches are disjoint, so “No Tree is a Branch.”
- Hence of the pair (I) “Some branches are trees” and (III) “No tree is a branch,” exactly one can be true; given the premises, (I) is false and (III) is true.
Step-by-Step Solution:
Derive (II): T ⊆ F and F ∩ Fr = ∅ ⇒ T ∩ Fr = ∅. So “No fruit is a tree” must be true.Assess (I) vs (III): B ⊆ Fr and T ∩ Fr = ∅ ⇒ T ∩ B = ∅, which makes (I) false and (III) true. The option format groups them as “either I or III” along with (II); effectively, the valid combination is (III) and (II).
Verification / Alternative check:
Any Venn diagram respecting the premises places Trees entirely away from Fruits (and thus away from Branches). This makes (II) and (III) inevitable and contradicts (I).
Why Other Options Are Wrong:
Claiming only (II) ignores the forced disjointness with Branches (supporting III). Claiming (I) contradicts the established disjointness.
Common Pitfalls:
Missing that “No Flower is a Fruit” propagates through the subset “All Trees are Flowers,” eliminating any Tree–Fruit overlap.
Final Answer:
Only either I or III, and II follow
Discussion & Comments